# data[].attributes.attachment_link.title
Related Rates # 7 - Ladder Sliding Down Wall, Finding Rate of Change of Area Under Ladder

# data[].attributes.attachment_link.readable_text
- [Voiceover] In the last couple videos, I showed how you can take a function, ah, just a function with two inputs, and find the tangent plane to its graph, and the way that you think about this, you first find a point, some kind of input point, which is, you know I'll just write abstractly as x nought and y nought. And you see where that point ends up on the graph, and you wanna find a new function, a new function which we were calling L, and maybe you say L sub f, which also is a function of x and y. And you want the graph of that function to be a plane tangent to the graph. Now this often goes by another name. This will go under the name Local Linearization, Local linearization, this is kind of a long word, zation. And what this basically means, the word local means you're looking at a specific input point. So in this case, it's a specific input point x nought, y nought, and the idea of a linearization, a linearization, means you're approximating the function with something simpler, with something that's actually linear, and I'll tell you what I mean by linear in, in just a moment. But the whole idea here is that we don't really care about, you know, tangent planes in an abstract 3D spaced to some kind of graph. The whole reason for doing this, is that this is a really good way to approximate a function, which is potentially a very complicated function with something that's much easier, something that has constant partial derivatives. Now my goal of this video is gonna be to show how we write this local linearization here in vector form, because it'll be both more compact, and hopefully easier to remember, and also it's more general. It'll apply to things that have more than just, just two input variables like this one does. So just to remind us of where we were, and what we got to in the last couple videos, I'll write a little bit more abstractly this time, rather than a specific example. The way you do this local linearization is first you find the partial derivative of f with respect to x, which I'll write with the subscript notation. And you evaluate that at x of o or x nought, y nought. You evaluate it at the point about which you're approximating and then you multiply that by x minus that constant. So the only variable right here, everything is a constant, but the only variable part is that x. And then we add to that, basically doing the same thing with y. You take the partial derivative with respect to y, you evaluate it at the input point, the point about which you are linearizing, and then you multiply it by y minus ys of o. And then to this entire thing because you wanna make sure that when you evaluate this function at the input point itself. You see, when you plug in x nought and y nought, this terms goes to zero, cause x nought minus x nought is zero. This terms goes to zero, and this is the whole reason we kind of paired up these terms and organized the constants in this way. This way, you can just think about adding whatever the function itself evaluates to at that point. And this will ensure that your linearization actually equals the function itself at the local point. Cause hopefully if you're approximating it near a point, then at that point, it's actually equal. So what do I mean by this word linear? The word linear has a very precise formulation, especially in the context of linear algebra, and admittedly, this is not actually a linear function in the technical sense. But loosely what it means, and the reason people call it linear, is that this x term here, this variable term, doesn't have anything fancy going on with it. It's just being multiplied by a constant, and similarly this y term it's just being multiplied by a constant. It's not squared, there's no square root, it's not in an exponent or anything like that. And although there is a more technical meaning of the word linear, this is all it really needs in this context. This is all you need to think about. Each variable is just multiplied by a constant. Now you might see this in a more complicated form, or what's at first a more complicated form using vectors. So first of all, let's think about how we would start describing everything going on here with vectors. So the input, ah, rather than talk about the input as being a pair of points, what I wanna say is that there's some vector, some vector that has these as its components, and we just wanna capture that all, and I wanna give that a name. And kind of unfortunately the name that we give this, it's very common to just call it x, and maybe a bold-faced x and that would be easier to do typing than it is writing, so I'll just kind of try to emphasize bold-faced x equals this vector, and where's, that's confusing cause x is already one of the input variables that's just a number. Um, but I'll try to emphasize it, just making it bold. You'll see this in writing a lot. X is this input vector, and then similarly, the specified, ah, specified input about which we are approximating, you would call, see I'll make it a nice bold-faced x nought. We'll, we'll do that nought to just kind of indicate that it's a constant of some kind. And what that is, it's a vector containing the two numbers x nought, y nought. So this is just us starting to write things in a more vectorized way and the convenience here is that, that if you're dealing with a function with three input variables or four or a hundred, you could still just write it as this bold-faced x with the understanding that the vector has a lot more components. So now, let's take a look at, at these first two terms in our linearization. We can start thinking of this as a, as a dot product, actually. So let me first just kind of move this guy out of the way and give ourselves some room. So he's gonna just go up there, this the same guy, and now I wanna think about writing this other term here as a, as a dot product. And what that looks like is we have the two partial derivatives fs of x and fs of y, indicating the partial derivatives with respect to x and y, and each one of them is evaluated. Let's see, I'll do it, I'll do it evaluating at our bold-faced x nought, and then this one is also evaluated at that bold-faced x nought. So, really you're thinking about this as being, you know, a vector that contains two different variables. You're just packing it into a single symbol, and the dot product here is against, ah, you know, the first component is x minus x nought, so I'd write that as x minus x nought the number, and then similarly, y minus, let's see, I'll do it in the same color, y nought the number. Ah, but we can write each one of these in a more compact form, where this, the vector that has the partial derivatives, that's the gradient, and if that feels unfamiliar, maybe go back and check out the videos on the gradient, but this whole vector is basically just saying, take the gradient and evaluate it at that, that vector input, you know, x nought. And in the second component here, that's telling you you've got x and y minus x nought and y nought. So what you're basically doing is taking the, you know, bold-faced input, the variable vector x, and then you're subtracting off, you know, x nought, where x nought is some kind of constant. So this right here, this is just vector terms where you're thinking of this as being a vector with two components, and this one is a vector with two components, but if your function happened to be something more complicated, with, you know, a hundred input variables, this would be the same thing you write down. You would just understand that when you expand this, there's gonna be a hundred different components in the vector. Um, and this is what a linear term looks like in vector terminology, cause this dot product is telling you that all of the components of that bold-faced x vector, the, that expands into, you know, not bold-faced x, y, z, whatever else it expands to. All of those are just being multiplied by some kind of constant. So we take that whole thing, that's, that's how you simplify the first couple terms here, and of course, we just add on the value of the function itself. So you would take that as the linear term. And no, I kind of like to add it on to the front, actually, where you think about taking the function itself and evaluating it at that, that constant input x nought, cause that way you can kind of think this is your constant term, this is your constant term, and then the rest of the stuff here is your linear term. Rest of your stuff is your linear. Cause later on if we start adding other terms like a quadratic term or more complicated things, you can kind of keep adding them on the end. So this right here, is the expression that you will often see for the local linearization. And the only place where the actual variable shows up, the variable vector, is right here, is this guy. Cause, you know, when you evaluate the function f at a specified input, that's just a constant. when you evaluate the gradient at that input, it's just a constant. And we're subtracting off that, that um, specified input that's just a constant. So this is the only place where your variable shows up. So once all is said and done, and once you do your computations, this is a very simple function. And the, the important part is maybe this is much simpler than the function f itself, which allows you to, you know, maybe compute something more quickly if you're writing a program that needs to, you know, deal with some kind of complicated function, but runtime is an issue or maybe, it's a function that you never knew in the first place, but you were able to approximate its value at a point, and approximate its gradient. So this is what lets you approximate the function as a whole near that point. So again, this might look very abstract, but if you just kind of unravel everything and think back to where it came from and look at, look at the specific example of a, you know, tangent plane, um, hopefully it all makes a little bit of sense and you see that this is really just the simplest possible function that evaluates to the same value as f when you input this point, and whose partial derivatives all evaluate to the same values as those of f at that specified point. And if you wanna see more examples of this, and what it looks like and maybe how you can use it to approximate certain functions, I have an article on that, that you can go check out, and it would be particularly good to kind of go in with a piece of paper and sort of work through the examples yourself as you, as you work through it. And with that said, I will see you next video.

# data[].attributes.attachment_link.title
Local linearization (video) | Khan Academy

# data[].attributes.attachment_link.description
Voiceover] In the last couple videos, I showed how you can take a function, ah, just a function with two inputs, and find the tangent plane to its graph, and the way that you think about this, you first find a point, some kind of input point, which is, you know I'll just write abstractly as x nought and y nought.

# data[].attributes.attachment_link.description
This lesson is about using the tangent line to approximate another point on our curve. What Is Linear Approximation The idea behind local linear

# data[].attributes.attachment_link.readable_text
Derivatives are all about change ...

... they show how fast something is changing (called the rate of change) at any point.

In Introduction to Derivatives (please read it first!) we looked at how to do a derivative using differences and limits.

Here we look at doing the same thing but using the "dy/dx" notation (also called Leibniz's notation) instead of limits.

We start by calling the function "y":

y = f(x)

1. Add Δx

When x increases by Δx, then y increases by Δy :

y + Δy = f(x + Δx)

2. Subtract the Two Formulas

3. Rate of Change

To work out how fast (called the rate of change) we divide by Δx:

4. Reduce Δx close to 0

We can't let Δx become 0 (because that would be dividing by 0), but we can make it head towards zero and call it "dx":

Δx dx

You can also think of "dx" as being infinitesimal, or infinitely small.

Likewise Δy becomes very small and we call it "dy", to give us:

Try It On A Function

Let's try f(x) = x ²

Why don't you try it on f(x) = x³ ?

What derivative do you get?

# data[].attributes.attachment_link.description
Derivatives are all about change ... ... they show how fast something is changing (called the rate of change) at any point. In Introduction to Derivatives (please read it first!) we looked at how to do a derivative using differences and limits.

# data[].attributes.attachment_link.readable_text
L'Hôpital's Rule can help us calculate a limit that may otherwise be hard or impossible.

L'Hôpital is pronounced "lopital". He was a French mathematician from the 1600s.

It says that the limit when we divide one function by another is the same after we take the derivative of each function (with some special conditions shown later).

In symbols we can write:
The limit as x approaches c of "f-of−x over g-of−x" equals the
the limit as x approaches c of "f-dash-of−x over g-dash-of−x"
All we did is add that little dash mark ' on each function, which means to take the derivative.

Example:

At x=2 we would normally get:

Which is indeterminate, so we are stuck. Or are we?

Let's try L'Hôpital!

Differentiate both top and bottom (see Derivative Rules):

Now we just substitute x=2 to get our answer:

Here is the graph, notice the "hole" at x=2:

Note: we can also get this answer by factoring, see Evaluating Limits.

Example:

Normally this is the result:

Both head to infinity. Which is indeterminate.

But let's differentiate both top and bottom (note that the derivative of e x is e x):

Hmmm, still not solved, both tending towards infinity. But we can use it again:

Now we have:

It has shown us that e x grows much faster than x ².

Cases

We have already seen a and example. Here are all the indeterminate forms that L'Hopital's Rule may be able to help with:

Conditions

Differentiable

For a limit approaching c, the original functions must be differentiable either side of c, but not necessarily at c.

Likewise g'(x) is not equal to zero either side of c.

The Limit Must Exist
This limit must exist:
Why? Well a good example is functions that never settle to a value.

Example:

Which is a case. Let's differentiate top and bottom:

And because it just wiggles up and down it never approaches any value.

So that new limit does not exist!
And so L'Hôpital's Rule is not usable in this case.
BUT we can do this:

And we are left with just the "1", so:

# data[].attributes.attachment_link.description
L'Hôpital's Rule can help us calculate a limit that may otherwise be hard or impossible. L'Hôpital is pronounced "lopital". He was a French mathematician from the 1600s. It says that the limit when we divide one function by another is the same after we take the derivative of each function (with some special conditions shown later).

# data[].attributes.headline
www.fiveable.com Study Guide for Unit 4 Topic 4.1-4.7

# data[].attributes.subject
www.fiveable.com Study Guide for Unit 4 Topic 4.1-4.7

# data[].attributes.attachment_link.title
2022 AP Calc Unit 4 Review | Free Reviews, Study Guides, & Notes | Fiveable

# data[].attributes.attachment_link.description
Fiveable is best place to study for your AP® exams. Free AP Calculus AB/BC study guides for Unit 4: Contextual Applications of the Differentiation. Join us on Discord!

# data[].attributes.attachment_link.readable_text
There are various applications of derivatives not only in maths and real life but also in other fields like science, engineering, physics, etc. In previous classes, you must have learned to find the derivative of different functions, like, trigonometric functions, implicit functions, logarithm functions, etc. In this section, you will learn the use of derivatives with respect to mathematical concepts and in real-life scenarios. This is one of the important topics covered in Class 12 Maths as well.

Derivatives have various important applications in Mathematics such as:

Rate of Change of a Quantity

Increasing and Decreasing Functions

Tangent and Normal to a Curve

Minimum and Maximum Values

Newton's Method

Linear Approximations

Applications of Derivatives in Maths

The derivative is defined as the rate of change of one quantity with respect to another. In terms of functions, the rate of change of function is defined as dy/dx = f(x) = y'.

The concept of derivatives has been used in small scale and large scale. The concept of derivatives used in many ways such as change of temperature or rate of change of shapes and sizes of an object depending on the conditions etc.,

Rate of Change of a Quantity

This is the general and most important application of derivative. For example, to check the rate of change of the volume of a cube with respect to its decreasing sides, we can use the derivative form as dy/dx. Where dy represents the rate of change of volume of cube and dx represents the change of sides of the cube.

Increasing and Decreasing Functions

To find that a given function is increasing or decreasing or constant, say in a graph, we use derivatives. If f is a function which is continuous in [p, q] and differentiable in the open interval (p, q), then,

f is increasing at [p, q] if f'(x) > 0 for each x ∈ (p, q)

f is decreasing at [p, q] if f'(x) < 0 for each x ∈ (p, q)

f is constant function in [p, q], if f'(x)=0 for each x ∈ (p, q)

Tangent and Normal To a Curve

Tangent is the line that touches the curve at a point and doesn't cross it, whereas normal is the perpendicular to that tangent.

Let the tangent meet the curve at P(x ₁, y ₁).

Now the straight-line equation which passes through a point having slope m could be written as;

We can see from the above equation, the slope of the tangent to the curve y = f(x) and at the point P(x ₁, y ₁), it is given as dy/dx at P(x ₁, y ₁) = f'(x). Therefore,

Equation of the tangent to the curve at P(x ₁, y ₁) can be written as:

Equation of normal to the curve is given by;

Or

Maxima and Minima

To calculate the highest and lowest point of the curve in a graph or to know its turning point, the derivative function is used.

When x = a, if f(x) ≤ f(a) for every x in the domain, then f(x) has an Absolute Maximum value and the point a is the point of the maximum value of f.

When x = a, if f(x) ≤ f(a) for every x in some open interval (p, q) then f(x) has a Relative Maximum value.

When x= a, if f(x) ≥ f(a) for every x in the domain then f(x) has an Absolute Minimum value and the point a is the point of the minimum value of f.

When x = a, if f(x) ≥ f(a) for every x in some open interval (p, q) then f(x) has a Relative Minimum value.

Monotonicity

Functions are said to be monotonic if they are either increasing or decreasing in their entire domain. f(x) = e x, f(x) = n x, f(x) = 2x + 3 are some examples.

Functions which are increasing and decreasing in their domain are said to be non-monotonic

For example: f(x) = sin x , f(x) = x ²
Monotonicity Of A function At A Point
A function is said to be monotonically decreasing at x = a if f(x) satisfy;

f(x + h) < f(a) for a small positive h

f'(x) will be positive if the function is increasing

f'(x) will be negative if the function is decreasing

f'(x) will be zero when the function is at its maxima or minima

Approximation or Finding Approximate Value

To find a very small change or variation of a quantity, we can use derivatives to give the approximate value of it. The approximate value is represented by delta △.

Suppose change in the value of x, dx = x then,

dy/dx = △x = x.

Since the change in x, dx ≈ x therefore, dy ≈ y.

Point of Inflection

For continuous function f(x), if f'(x ₀) = 0 or f'"(x ₀) does not exist at points where f'(x ₀) exists and if f"(x) changes sign when passing through x = x 0 then x ₀ is called the point of inflection.

(a) If f"(x) < 0, x ∈ (a, b) then the curve y = f(x) in concave downward

(b) if f" (x) > 0, x ∈ (a, b) then the curve y = f(x) is concave upwards in (a, b)
For example: f(x) = sin x
Solution: f'(x) = cos x

f"(x) = sinx = 0 x = nπ, n ∈ z

Application of Derivatives in Real Life

To calculate the profit and loss in business using graphs.

To check the temperature variation.

To determine the speed or distance covered such as miles per hour, kilometre per hour etc.

Derivatives are used to derive many equations in Physics.

In the study of Seismology like to find the range of magnitudes of the earthquake.

By solving the application of derivatives problems, the concepts for these applications will be understood in a better manner.
Also, read:
Applications of Derivatives Questions
Q.1: Show that the function f(x) = x³ - 2x2 + 2x, x ∈ Q is increasing on Q.
By differentiating both sides, we get,

f'(x) = 3x 2 - 4x + 2 > 0 for every value of x

Therefore, f is increasing on Q.
Q.2: The tangent to the curve y=x2 − 5x + 5 parallel to the line 2y = 4x + 1, also passes through a point. Find the coordinates of the point.
Solution:

dy/dx = 2x - 5| x=x1 = 2

y ₁ = (49/4) - (35/2) + 5 = (49 - 70 + 20)/4 = -¼

y + ¼ = 2(x - 7/2)

4y + 1 = 8x - 28

⇒ 8x - 4y - 29 = 0

x = ⅛, y = -7 satisfies the equation
Q.3: The tangent to the curve, \(y=x{{e}^{{{x}^{2}}}}\) passing through the point (1, e) also passes through another point. Find it.
Solution: \(\frac{dy}{dx}={{e}^{{{x}^{2}}}}+x{{e}^{{{x}^{2}}}}.\,2x\)

At x = 1, slope of tangent m = 3e

Equation of tangent :

y−e = 3e(x−1)

⇒y = 3ex − 2e

(4/3,2e) lies on it.

Learn more important maths fundamentals and topics by downloading BYJU'S- The learning App.

# data[].attributes.attachment_link.title
Applications Of Derivatives in Maths and in Real Life (With Examples)

# data[].attributes.attachment_link.description
There are various applications of derivatives not only in maths and real life but also in other fields like science, engineering, physics, etc. In previous classes, you must have learned to find the derivative of different functions, like, trigonometric functions, implicit functions, logarithm functions, etc.

# data[].attributes.attachment_link.readable_text
Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax.

Attribution information

If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction

If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction

Citation information
Use the information below to generate a citation. We recommend using a citation tool such as this one.

© Jun 15, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

# data[].attributes.attachment_link.title
Ch. 4 Introduction - Calculus Volume 1 | OpenStax

# data[].attributes.attachment_link.description
Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax.

# data[].attributes.headline
www.ixl.com Pre-Cal: CC Introduction to Derivatives

# data[].attributes.subject
www.ixl.com Pre-Cal: CC Introduction to Derivatives

# data[].attributes.attachment_link.description
Set students up for success in Precalculus and beyond! Explore the entire Precalculus curriculum: polynomials, derivatives, and more. Try it free!

# data[].attributes.attachment_link.title
AP Calculus AB and BC Unit 4 Review [Contextual Applications of Differentiation]

# data[].attributes.headline
Playlist for Khan Academy: Motion along a line

# data[].attributes.subject
Playlist for Khan Academy: Motion along a line

# data[].attributes.attachment_link.readable_text
Let's say that we have some particle that's moving along the number line. So let me draw a number line right over here. So that's our number line right over there. And let's say it starts right over here at 0. And then as time passes, this little point is going to move around. Maybe it moves to the right, slows down, speeds up. Maybe it moves to the left, slows down, speeds up. It might do all sorts of things. And to describe this motion, its position as a function of time, we have a function s of t. This particle's position as a function of time we're given is t to the third power minus 6t t squared plus 9t. And we're going to restrict the domain to positive time. So we're going to assume that time is greater than or equal to 0. Now the question that we want to answer in this video is, when is this particle speeding up? So when are we speeding up? And I think that bears a little clarification. What does it mean to speed up? Well, there's two scenarios. If the particle is already moving in the rightward direction-- and the way we would know it's moving in the rightward direction is if its velocity is greater than 0. If it's moving in the rightward direction and it's also accelerating in the rightward direction-- so if its acceleration is also greater than 0-- then this is a situation where we are speeding up. Now another scenario where we would be speeding up is if we're moving in the leftward direction. In that case, our velocity is going to be negative. So if our velocity is negative and we want to go faster in the negative direction, then our acceleration should also be negative. That would make our velocity getting more and more and more negative with time. So then our acceleration needs to also be negative if we still want to be speeding up. If you have any other combination here, if your velocity is negative but your acceleration is positive, that means that your velocity is becoming less negative, or you would be slowing down the leftward direction. And vice versa, if your velocity is positive and your acceleration is negative, that means you're going to the right but you are slowing down in the rightward direction. So let's think about these two scenarios. And since velocity matters here so much, we just have to remind ourselves that the velocity-- remember, a derivative is just the rate of change with respect to a variable. So if you have your position function, the derivative of position with respect to time, this is really just what is the instantaneous rate of change of position with respect to time? Well, what is the change of position with respect to time? Well that is just going to be equal to our velocity function. That's going to be equal to our velocity function, v of t. Or we could write s prime of t, which could be also written this way, as ds dt, is equal to our velocity as a function of time. So let's take the derivative of this. Our velocity as a function of time is going to be equal to 3t squared minus 12t plus 9. So let's see if we can graph this velocity function to start making sense of it. When is the velocity positive? When is it negative? And what's the acceleration doing in those intervals? And so to help me graph it, we could say the v-intercept, or the vertical intercept, when v of 0 is going to be equal to 9. So that'll help us graph it. That's where we intersect the vertical axis. But also, let's plot-- let's figure out where it intersects the t-axis. So let's set this equal to 0. So 3t squared minus 12t plus 9 is equal to 0. Let's see. To simplify this, I can divide both sides by 3. And I get t squared minus 4t plus 3 is equal to 0. Now this is very factorable. This is t. Let's see. What two numbers, when you take a product, get 3, and when you add them, you get negative 4? Well, that's going to be t minus 3 times t minus 1 is equal to 0. How can this expression be equal to 0? Well if either of these are equal to 0, if either t minus 3 is 0 or t minus 1 is 0, it's going to be equal to 0. So t could be equal to 3, or t could be equal to 1. If t is 3 or t is 1, either of these are equal to 0, or this entire expression up here is going to be equal to 0. And since our coefficient on the t squared term is positive, we know this is going to be an upward opening parabola. So let's see if we can plot velocity as a function of time. So that is my velocity axis. This right over here is my time axis. And let's say this is 1 times 1 second, or I'm assuming this is in seconds-- 2, 3, 4. Actually, let me spread them apart a little bit more just because 1 and 3 are significant-- 1, 2, and 3. And they're not going to be-- I'm going to squash to the vertical scale a little bit. But this right over here, let's say that is 9, a velocity of 9. And so when t equals 0, our velocity is 9. When t equals 1, then our velocity is going to be 0. We get that right over here. 3 minus 12 plus 9, that's 0. And when t is equal to 3 our velocity is 0 again. Our vertex is going to be right in between those, when t is equal to 2-- right in between these two 0's. And we could figure out what that velocity is if we like. It's going to be 3 times 4 minus 12 times 2 plus 9. So what is that? That's 12 minus 24 plus 9. So that is negative 12 plus 9. So that's going to be equal to negative 3. Did I do that-- 12, yep, negative 3. So you're going to be-- negative 3 might be-- that's 9, so that's positive. So it might be something like this. So the graph of our velocity as a function of time is going to look something like this. And we only care about positive time. It's going to look something like this. So let's think. Remember, this is velocity. This is our velocity as a function of time. Now let's think about when is the velocity less than 0 and the acceleration is less than 0? So let's think about this case right over here? When is this the case? Both of them are going to be less than 0. Well, velocity is the less than 0 over this entire interval, this entire magenta interval. But the acceleration isn't less than 0 that entire time. Remember, the acceleration is the rate of change of velocity. We can write here that acceleration as a function of time, this is equal to the rate which velocity changes with respect to time. Or we could write, acceleration is equal to v prime of t, which is the same thing as the second derivative of position with respect to time. And so the acceleration, you could really think of the slope of the tangent line of the velocity function. And so over here, the place where this is downward sloping, where this has a negative slope, and the curve itself is below the t-axis, that's only over this interval right over here. Between this 0 right over here and the vertex, we get to this point right over here. And then our slope flattens out. So this interval right over here is t is going to be greater than 1, and it is going to be less than 2. That meets these constraints. Now let's think about where our velocity is greater than 0 and our acceleration is greater than 0. Well our velocity is greater than 0 over here. But notice, our acceleration, the slope here is negative. We're downward sloping, so that doesn't apply. Here our velocity is greater than 0, and the slope of the velocity, the rate of change of velocity, the acceleration, is also greater than 0. So that's this interval right over here, where we're speeding up in the rightward direction. So that interval is t is greater than 3. So when are we speeding up? We're speeding up between the first and second seconds, and then we're speeding up after the third second.

# data[].attributes.attachment_link.title
Motion problems: when a particle is speeding up (video) | Khan Academy

# data[].attributes.attachment_link.description
Let's say that we have some particle that's moving along the number line. So let me draw a number line right over here. So that's our number line right over there. And let's say it starts right over here at 0. And then as time passes, this little point is going to move around.

# data[].attributes.headline
Playlist for Khan Academy: Applied Rates of Change

# data[].attributes.subject
Playlist for Khan Academy: Applied Rates of Change

# data[].attributes.attachment_link.readable_text
I studied for an English test today and learned 80 vocabulary words. In 10 days, I will have forgotten every word. The number of words that I remember t days after studying is modeled by-- so W of t, so this is the number of words I have in my head as a function of time is going to be equal to 80 times 1 minus 0.1t squared for t is between 0 and 10, including the two boundaries. That's why we have brackets right over here. What is the rate of change of the number of known words per day two days after studying for the test? And I encourage you to pause this video and try it on your own. So the key here is we come up with this equation for modeling how many words have retained in my brain every day after I first memorized them, after I got the 80 of them into my head? And that's this expression here. And they want to know the rate of change two days after studying. Well, the rate of change, I can take the derivative of this with respect to time. So let's do that. So let's take the derivative. The derivative of the number of words I know with respect to time is going to be equal to-- well, we have this 80 out front. That's just a constant. And now I can apply the chain rule right over here. So the derivative of 1 minus 0.1t, the whole thing squared with respect to 1 minus 0.1t is going to be-- so I'm essentially taking the derivative of this whole pink thing, this whole expression squared with respect to the expression. So that's going to be 2 times 1 minus 0.1t. And now I can find the derivative of this inner expression with respect to t. So the derivative of this inner expression with respect to t is just going to be 0 minus 0.1. So it's just going to be negative 0.1. And of course, we can simplify this a little bit. This is going to be equal to-- if we take 80 times 2 is 160 times negative 0.1, that's going to be negative 16. 160 times 0.1 is 16, so negative 16 times 1 minus 0.1t. And if we want, we could distribute the 16, or we could just leave it like this. But we're ready now to answer our question. We could write this as the rate of change of the number of words we know with respect to time. Or we could use the alternate notation. We could say this is W prime of t. Either way, it's going to be equal to this thing. Let me do that same color. It's equal to negative 16 times 1 minus 0.1t. So what's this going to be? What is the rate of change of the number of words known per day two days after studying for the test? Well, we just have to evaluate this when t is equal to 2. So W prime of 2 is going to be equal to negative 16 times 1 minus 0.1 times 2 close parentheses. And that's going to be equal to-- well, let's see, what is this? This is 1 minus essentially 0.2. This is going to be 0.8. So this is going to be equal to negative 16 times-- is that right? 1 minus 0.2 is-- yep, it's going to be times 0.8. And what is that going to be? If I were to multiply 16 times 8, it would be 128. It's 2 times 8 times 8 so 2 times 64, 128. So this is going to be negative 12.8. And just to really hit the point home of what we're doing, the rate of change is negative 12.8 words per day. So if you believe this model for how many words we know on a given day, this is saying on day two, right at the day two point, right after exactly two days after studying for the test, right at that moment, I am essentially losing 12.8 words per day. The number of words I know is decreasing by 12.8 words per day.

# data[].attributes.attachment_link.title
Applied rate of change: forgetfulness (video) | Khan Academy

# data[].attributes.attachment_link.description
I studied for an English test today and learned 80 vocabulary words. In 10 days, I will have forgotten every word. The number of words that I remember t days after studying is modeled by-- so W of t, so this is the number of words I have in my head as a function of time is going to be equal to 80 times 1 minus 0.1t squared for t is between 0 and 10, including the two boundaries.

# data[].attributes.attachment_link.readable_text
You appear to be on a device with a "narrow" screen width ( i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

1. \(\displaystyle {\left( {f\left( x \right) \pm g\left( x \right)} \right)^\prime } = f'\left( x \right) \pm g'\left( x \right)\hspace{0.25in} \mbox{OR} \hspace{0.25in}\frac{d}{{dx}}\left( {f\left( x \right) \pm g\left( x \right)} \right) = \frac{{df}}{{dx}} \pm \frac{{dg}}{{dx}}\)
In other words, to differentiate a sum or difference all we need to do is differentiate the individual terms and then put them back together with the appropriate signs. Note as well that this property is not limited to two functions.

See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this property. It's a very simple proof using the definition of the derivative.

2. \(\displaystyle {\left( {cf\left( x \right)} \right)^\prime } = cf'\left( x \right)\hspace{0.25in} \mbox{OR} \hspace{0.25in}\frac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\frac{{df}}{{dx}}\), \(c\) is any number
In other words, we can "factor" a multiplicative constant out of a derivative if we need to. See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this property.

1. If \(f\left( x \right) = c\) then \(\displaystyle f'\left( x \right) = 0\hspace{0.25in} \mbox{OR} \hspace{0.25in}\frac{d}{{dx}}\left( c \right) = 0\)
The derivative of a constant is zero. See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this formula.

2. If \(f\left( x \right) = {x^n}\) then \(\displaystyle f'\left( x \right) = n{x^{n - 1}}\hspace{0.25in} \mbox{OR} \hspace{0.25in}\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\), \(n\) is any number.
This formula is sometimes called the power rule. All we are doing here is bringing the original exponent down in front and multiplying and then subtracting one from the original exponent.

Note as well that in order to use this formula \(n\) must be a number, it can't be a variable. Also note that the base, the \(x\), must be a variable, it can't be a number. It will be tempting in some later sections to misuse the Power Rule when we run in some functions where the exponent isn't a number and/or the base isn't a variable.

See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof of this formula. There are actually three different proofs in this section. The first two restrict the formula to \(n\) being an integer because at this point that is all that we can do at this point. The third proof is for the general rule but does suppose that you've read most of this chapter.

Example 1 Differentiate each of the following functions.

1. \(f\left( x \right) = 15{x^{100}} - 3{x^{12}} + 5x - 46\)

2. \(g\left( t \right) = 2{t^6} + 7{t^{ - 6}}\)

3. \(y = 8{z^3} - \frac{1}{{3{z^5}}} + z - 23\)

4. \(\displaystyle T\left( x \right) = \sqrt x + 9\sqrt[3]{{{x^7}}} - \frac{2}{{\sqrt[5]{{{x^2}}}}}\)

5. \(h\left( x \right) = {x^\pi } - {x^{\sqrt 2 }}\)

Show All Solutions Hide All Solutions
a \(f\left( x \right) = 15{x^{100}} - 3{x^{12}} + 5x - 46\) Show Solution

In this case we have the sum and difference of four terms and so we will differentiate each of the terms using the first property from above and then put them back together with the proper sign. Also, for each term with a multiplicative constant remember that all we need to do is "factor" the constant out (using the second property) and then do the derivative.

\[\begin{align*}f'\left( x \right) & = 15\left( {100} \right){x^{99}} - 3\left( {12} \right){x^{11}} + 5\left( 1 \right){x^0} - 0\\ & = 1500{x^{99}} - 36{x^{11}} + 5\end{align*}\]

Notice that in the third term the exponent was a one and so upon subtracting 1 from the original exponent we get a new exponent of zero. Now recall that \({x^0} = 1\). Don't forget to do any basic arithmetic that needs to be done such as any multiplication and/or division in the coefficients.

b \(g\left( t \right) = 2{t^6} + 7{t^{ - 6}}\) Show Solution

The point of this problem is to make sure that you deal with negative exponents correctly. Here is the derivative.

\[\begin{align*}g'\left( t \right) & = 2\left( 6 \right){t^5} + 7\left( { - 6} \right){t^{ - 7}}\\ & = 12{t^5} - 42{t^{ - 7}}\end{align*}\]

Make sure that you correctly deal with the exponents in these cases, especially the negative exponents. It is an easy mistake to "go the other way" when subtracting one off from a negative exponent and get \( - 6{t^{ - 5}}\) instead of the correct \( - 6{t^{ - 7}}\).

Now in this function the second term is not correctly set up for us to use the power rule. The power rule requires that the term be a variable to a power only and the term must be in the numerator. So, prior to differentiating we first need to rewrite the second term into a form that we can deal with.

\[y = 8{z^3} - \frac{1}{3}{z^{ - 5}} + z - 23\]

Note that we left the 3 in the denominator and only moved the variable up to the numerator. Remember that the only thing that gets an exponent is the term that is immediately to the left of the exponent. If we'd wanted the three to come up as well we'd have written,

\[\frac{1}{{{{\left( {3z} \right)}^5}}}\]

so be careful with this! It's a very common mistake to bring the 3 up into the numerator as well at this stage.

Now that we've gotten the function rewritten into a proper form that allows us to use the Power Rule we can differentiate the function. Here is the derivative for this part.

\[y' = 24{z^2} + \frac{5}{3}{z^{ - 6}} + 1\]

d \(\displaystyle T\left( x \right) = \sqrt x + 9\sqrt[3]{{{x^7}}} - \frac{2}{{\sqrt[5]{{{x^2}}}}}\) Show Solution

All of the terms in this function have roots in them. In order to use the power rule we need to first convert all the roots to fractional exponents. Again, remember that the Power Rule requires us to have a variable to a number and that it must be in the numerator of the term. Here is the function written in "proper" form.

\[\begin{align*}T\left( x \right) & = {x^{\frac{1}{2}}} + 9{\left( {{x^7}} \right)^{\frac{1}{3}}} - \frac{2}{{{{\left( {{x^2}} \right)}^{\frac{1}{5}}}}}\\ & = {x^{\frac{1}{2}}} + 9{x^{\frac{7}{3}}} - \frac{2}{{{x^{\frac{2}{5}}}}}\\ & = {x^{\frac{1}{2}}} + 9{x^{\frac{7}{3}}} - 2{x^{ - \frac{2}{5}}}\end{align*}\]

In the last two terms we combined the exponents. You should always do this with this kind of term. In a later section we will learn of a technique that would allow us to differentiate this term without combining exponents, however it will take significantly more work to do. Also, don't forget to move the term in the denominator of the third term up to the numerator. We can now differentiate the function.

\[\begin{align*}T'\left( x \right) & = \frac{1}{2}{x^{ - \,\,\frac{1}{2}}} + 9\left( {\frac{7}{3}} \right){x^{\frac{4}{3}}} - 2\left( { - \frac{2}{5}} \right){x^{ - \,\,\frac{7}{5}}}\\ & = \frac{1}{2}{x^{ - \,\,\frac{1}{2}}} + \frac{{63}}{3}{x^{\frac{4}{3}}} + \frac{4}{5}{x^{ - \,\,\frac{7}{5}}}\end{align*}\]

Make sure that you can deal with fractional exponents. You will see a lot of them in this class.

e \(h\left( x \right) = {x^\pi } - {x^{\sqrt 2 }}\) Show Solution

In all of the previous examples the exponents have been nice integers or fractions. That is usually what we'll see in this class. However, the exponent only needs to be a number so don't get excited about problems like this one. They work exactly the same.

\[h'\left( x \right) = \pi {x^{\pi - 1}} - \sqrt 2 {x^{\sqrt 2 - 1}}\]

The answer is a little messy and we won't reduce the exponents down to decimals. However, this problem is not terribly difficult it just looks that way initially.

Example 2 Differentiate each of the following functions.

1. \(y = \sqrt[3]{{{x^2}}}\left( {2x - {x^2}} \right)\)

2. \(\displaystyle h\left( t \right) = \frac{{2{t^5} + {t^2} - 5}}{{{t^2}}}\)

Show All Solutions Hide All Solutions
a \(y = \sqrt[3]{{{x^2}}}\left( {2x - {x^2}} \right)\) Show Solution

In this function we can't just differentiate the first term, differentiate the second term and then multiply the two back together. That just won't work. We will discuss this in detail in the next section so if you're not sure you believe that hold on for a bit and we'll be looking at that soon as well as showing you an example of why it won't work.

It is still possible to do this derivative however. All that we need to do is convert the radical to fractional exponents (as we should anyway) and then multiply this through the parenthesis.

\[y = {x^{\frac{2}{3}}}\left( {2x - {x^2}} \right) = 2{x^{\frac{5}{3}}} - {x^{\frac{8}{3}}}\]

Now we can differentiate the function.

\[y' = \frac{{10}}{3}{x^{\frac{2}{3}}} - \frac{8}{3}{x^{\frac{5}{3}}}\]

b \(\displaystyle h\left( t \right) = \frac{{2{t^5} + {t^2} - 5}}{{{t^2}}}\) Show Solution

As with the first part we can't just differentiate the numerator and the denominator and the put it back together as a fraction. Again, if you're not sure you believe this hold on until the next section and we'll take a more detailed look at this.

We can simplify this rational expression however as follows.

\[h\left( t \right) = \frac{{2{t^5}}}{{{t^2}}} + \frac{{{t^2}}}{{{t^2}}} - \frac{5}{{{t^2}}} = 2{t^3} + 1 - 5{t^{ - 2}}\]

This is a function that we can differentiate.

\[h'\left( t \right) = 6{t^2} + 10{t^{ - 3}}\]

Example 3 Is \(\displaystyle f\left( x \right) = 2{x^3} + \frac{{300}}{{{x^3}}} + 4\) increasing, decreasing or not changing at \(x = - 2\)?

Show Solution

We know that the rate of change of a function is given by the functions derivative so all we need to do is it rewrite the function (to deal with the second term) and then take the derivative.

\[f\left( x \right) = 2{x^3} + 300{x^{ - 3}} + 4\hspace{0.25in} \Rightarrow \hspace{0.25in}f'\left( x \right) = 6{x^2} - 900{x^{ - 4}} = 6{x^2} - \frac{{900}}{{{x^4}}}\]

Note that we rewrote the last term in the derivative back as a fraction. This is not something we've done to this point and is only being done here to help with the evaluation in the next step. It's often easier to do the evaluation with positive exponents.

So, upon evaluating the derivative we get

\[f'\left( { - 2} \right) = 6\left( 4 \right) - \frac{{900}}{{16}} = - \frac{{129}}{4} = - 32.25\]

So, at \(x = - 2\) the derivative is negative and so the function is decreasing at \(x = - 2\).

Example 4 Find the equation of the tangent line to \(f\left( x \right) = 4x - 8\sqrt x \) at \(x = 16\).

Show Solution

We know that the equation of a tangent line is given by,

\[y = f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\]

So, we will need the derivative of the function (don't forget to get rid of the radical).

\[f\left( x \right) = 4x - 8{x^{\frac{1}{2}}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}f'\left( x \right) = 4 - 4{x^{ - \frac{1}{2}}} = 4 - \frac{4}{{{x^{\frac{1}{2}}}}}\]

Again, notice that we eliminated the negative exponent in the derivative solely for the sake of the evaluation. All we need to do then is evaluate the function and the derivative at the point in question, \(x = 16\).

\[f\left( {16} \right) = 64 - 8\left( 4 \right) = 32\hspace{0.25in}\hspace{0.25in}f'\left( 16 \right) = 4 - \frac{4}{4} = 3\]

The tangent line is then,

\[y = 32 + 3\left( {x - 16} \right) = 3x - 16\]

Example 5 The position of an object at any time \(t\) (in hours) is given by, \[s\left( t \right) = 2{t^3} - 21{t^2} + 60t - 10\]

Determine when the object is moving to the right and when the object is moving to the left.

Show Solution

The only way that we'll know for sure which direction the object is moving is to have the velocity in hand. Recall that if the velocity is positive the object is moving off to the right and if the velocity is negative then the object is moving to the left.

We need the derivative in order to get the velocity of the object. The derivative, and hence the velocity, is,

\[s'\left( t \right) = 6{t^2} - 42t + 60 = 6\left( {{t^2} - 7t + 10} \right) = 6\left( {t - 2} \right)\left( {t - 5} \right)\]

The reason for factoring the derivative will be apparent shortly.

Now, we need to determine where the derivative is positive and where the derivative is negative. There are several ways to do this. The method that we tend to prefer is the following.

Since polynomials are continuous we know from the Intermediate Value Theorem that if the polynomial ever changes sign then it must have first gone through zero. So, if we knew where the derivative was zero we would know the only points where the derivative might change sign.

We can see from the factored form of the derivative that the derivative will be zero at \(t = 2\) and \(t = 5\). Let's graph these points on a number line.

Now, we can see that these two points divide the number line into three distinct regions. In each of these regions we know that the derivative will be the same sign. Recall the derivative can only change sign at the two points that are used to divide the number line up into the regions.

Therefore, all that we need to do is to check the derivative at a test point in each region and the derivative in that region will have the same sign as the test point. Here is the number line with the test points and results shown.

Here are the intervals in which the derivative is positive and negative.

\[\begin{array}{rl}{{\mbox{positive : }}}&{ - \infty < t < 2\,\,\,\,\& \,\,\,\,5 < t < \infty }\\{{\mbox{negative : }}}&{2 < t < 5}\end{array}\]

We included negative \(t\)'s here because we could even though they may not make much sense for this problem. Once we know this we also can answer the question. The object is moving to the right and left in the following intervals.

\[\begin{array}{rl}{{\mbox{moving to the right : }}}&{ - \infty < t < 2\,\,\,\,\& \,\,\,\,5 < t < \infty }\\{{\mbox{moving to the left : }}}&{2 < t < 5}\end{array}\]

# data[].attributes.attachment_link.description
You appear to be on a device with a "narrow" screen width ( i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode.

# data[].attributes.attachment_link.title
Understanding the Definition of the Derivative