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The second component of the AP Exam includes 6 Free Response Questions (FRQs).

You're given 30 minutes to complete the first two-Part A. These both require a calculator. Then you'll have 1 hour for the remaining four-Part B. These, however, are tech-free. Not to worry, though, in the sections ahead we'll detail what you can expect in each part and how to maximize your score.

Speaking of scoring, the AP weights each problem equally, 9 points apiece (more later). And, taken together, this section accounts for 50% of your total exam score.

The AP course emphasizes 3 big ideas: change, limits, and analysis of functions. While the specific content varies from year-to-year, in aggregate the exam will test your ability to understand and interpret functions represented in a variety of forms: graphically, from a table, in written form, and as an equation.

Each FRQ contains 3 or 4 related sub-tasks, and you can expect a fairly even split between process-oriented and conceptual questions.

Process-Oriented: find the derivative; estimate the value of a definite integral; or identify the location or value of a relative maximum

Conceptual: interpret the graph of the first derivative, f'; verify the conditions of the Mean Value Theorem (MVT); explain why the rate of change of a system is positive or negative with justification

Note: these lists are not exhaustive, but the FAQs section below discusses other examples

In addition, two questions (at minimum) will feature an authentic scenario. This means they'll give you more context than an abstract equation or graph.

So what do you need to do differently in the FRQ section? In the words of every math teacher you've ever had: show your work.

This emphasis on justification is more than mere suggestion. Approximately half of the points possible depend on your ability to support your answer. Of course, what constitutes good justification depends on the question. See the comparisons below.

While this is accurate, it isn't clear that this is the derivative nor how we arrived here. A better answer would include evidence of our process.

Here, we clearly show we're taking the derivative-using the power rule and the chain rule. If we make an arithmetic error, the reader knows what we were attempting. In these situations you may earn yourself some of the available points. However, if you make a calculation error in the first solution, you'll earn no points at all.

Again, technically the result is accurate. Still, we could do much better to communicate our line of thinking.

Step one indicates that we've taken the derivative at x=3. Next we show the location of the point of intersection (3, p(3)). Finally, we label our result.

Before we move on, just a few more quick examples.

*NOTE: simply providing a +/- chart (a common practice for these types of problems) does not supply sufficient justification. A simple sentence indicating a change, however, will suffice.

As mentioned earlier, each FRQ counts equally. Within each sub-question, however the weightings can vary slightly. That said, normally each part counts 2 or 3 points toward the 9 point total. Thus, the maximum possible tally for the entire Free Response Section is 54 points. For reference, the MC section contributes 54 points, as well. Your raw score will sum these two.

If you're aiming for a 5? The cutoff changes every year, but a general guide is that raw scores above 70 should qualify.

And what about a 4? Here, again, it varies year-to-year. Usually, raw scores between 60-69 will translate to a 4.

AP readers-those people you're trying to please with your answers-award points based on accuracy and on completeness. To earn maximum marks you'll need both.

Above, we discussed some particulars of how to justify your answers. It's important because they contribute significantly to your total score.

That said, you will earn points for other things. You'll earn points for setting up derivatives and integrals properly; using complete and proper notation; logically moving step-by-step through your mathematical process; and providing the correct answer with the corresponding units, where appropriate. So, what does this look like?

Rates: At t = 3, the rate at which people are leaving is 142 every hour.

Average: On [3, 7] the average rate of change of v is 6 meters/sec.

Accumulation: The liquid has cooled 8 degrees Fahrenheit over the interval from t = 0 to t = 15 mins.

Exactly what do they mean when they say calculator required?

Well, you can expect to be asked to graph gnarly functions. They often won't give you a useful window, so you'll also need to find that. In these scenarios, try the ZoomFit option. It isn't always perfect, but it can be a lifesaver and (at minimum) point you in the right direction. This is usually just a first step.

Often the questions ask you to find zeros (keep at least 3 decimal places) or locate relative (or absolute) minimums and maximums.

Remember, if they ask for location, they're looking for the x-coordinate.

And if they ask for the absolute maximum? Then they want the y-value.

What else is your calculator good for? Calculating! The AP also may ask you to compute the value of a first or second derivative at a specific point. Similarly, they could ask for the numerical value of any type of definite integral. Need to practice? We've got you covered.

OK. But how am I supposed to justify? I'm glad you asked. Let's look at an example.

A local citrus vendor stocks her stand with 100 pounds of clementines at 9 a.m, at which time the market opens. She leaves her partner in control of the stand and returns to the farm to pick up more fruit. Customers buy her fruit at a rate of

where p(t) is in pounds and t is the number of hours since 9 a.m. Three hours later, the vendor returns with a resupply and moves them from the truck to the stand at a rate modeled by

And there you have it! In Part A, you'll lean heavily on your calculator. And as long as you provide good, clear mathematics notation, appropriate reasoning, and correct units, you'll be prepared to collect maximum points!

Many AP Calculus teachers attempt to predict the types of FRQs that will appear on the upcoming exam. As you'll see below, Modeling Rates and Particle Motion have been popular in the past few years.
2019: Modeling Rates and Particle Motion

1-integrate a difference of rates (definite); average value calculation (not average rate of change); find maximum; interpret rate of change (of a rate)

2-from a table, justify intermediate value theorem (IVT); approximate using trapezoidal rule; compute distance traveled given velocity; calculate a relative distance between two moving bodies

2018: Modeling Rates and Particle Motion

1-definite integral given a rate and bounds; integral evaluation with constant function adjustment; finding zeros; minimize a difference of functions

2-find and calculate derivative; evaluate a definite integral with initial condition; interpret displacement vs. total distance traveled

2017: Modeling Volume/Related Rates and Modeling Rates

1-Riemann sums approximation from a table; determine over/underestimate with justification; modeling volume from an equation given cross-sectional heights; applying the fundamental theorem of calculus (FTC)

2-compute a definite integral given a rate and bounds; calculate and interpret the derivative of a rate; find and explain the difference of two rates; compute a definite integral with an initial condition

Above all, whichever calculator you select, make sure you're familiar with it. It's useful to refresh (and practice!) the calculator functions you'll need on exam day.

In short, you'll need a graphing calculator for the exam. And while you've likely used one all year that falls within the guidelines, the AP is quite particular about which graphing calculator model you may use.

You are permitted to bring a backup calculator, though if you're using fresh batteries and pack a few extras (just in case) you should have plenty of insurance.

TI-73; T1-80; TI-81; TI-82; TI-83 (Plus and Plus Silver)

TI-84 (Plus/CE, Plus Silver, Plus C Silver, Plus T, Plus CE-T)

TI-85; TI-86; TI-89 and Titanium

TI-NSpire (CX, CX II, CX II-T)

TI-NSpire (CX II CAS, CX II-T CAS, CX II-C CAS)

FX-6000 Series; FX-6200 Series; FX-6300 Series

FX-6500 Series; FX-7000 Series; FX-7300 Series

FX-7400 Series; FX-7500 Series; FX-7700 Series

FX-7800 Series; FX-8000 Series; FX-8500 Series

FX-8700 Series; FX-8800 Series; FX-9700 Series

FX-9750 Series; FX-9860 Series; FX-CG-5000 (No Stylus)

FX 1.0 Series; FX-CG-10; FX-CG-20 Series; FX-CG-50

CFX-9800 Series; CFX-9850 Series; CFX-9950 Series

CFX-9970 Series; Algebra FX 2.0 Series

Graph25 Series; Graph35 Series; Graph75 Series

Graph95 Series; Graph100 Series

HP-9G; HP-28 Series; HP-38G; HP-39 Series; HP-40 Series HP-48 Series; HP-49 Series; HP-50 Series; HP Prime

4 Questions; 60 minutes; Graphing Calculator Prohibited
Alas, you'll need to put away your trusty TI (or other approved model) in the final section of the exam. Looking for consolation? Well, on a number of the problems your calculator wouldn't offer much help or time-savings anyway. How could that be?

Here the AP likes to test your ability to interpret graphs, tables, and slope fields.

What can you discern about g''(t)at t=a if you're given a picture of g(t)?

Use a table of f(t)values to guarantee the existence of a particular f'(t) value.

Could you produce a specific solution to a differential equation from a slope field?

These examples should give you some idea, but for clarity let's work an example.

The function g is differentiable on the closed interval [-6, 3]and satisfies g(-2) = 15. The graph of g' consists of a semicircle, a line segment, and a quarter-circle, as shown above, where g'(2) = 1.

Congratulations! Once you complete Part B, you've completed your exam. On test day remember to use good, clear mathematics notation, and always justify your answer!

As mentioned earlier, it's a popular pastime of some AP Calculus teachers to predict the content that will appear in the upcoming FRQs. You'll notice that a problem requiring you to analyze a graph and apply FTC has appeared with some frequency.
2019

Graphical Analysis of f and application of FTC

Modeling using Separable Differential Equations

Calculating Area and Volumes of Revolution

Function Analysis-L'Hospital and Squeeze

2018

Graphical Analysis of f' and application of FTC

Modeling from a table of data and analytically

Function Analysis-L'Hospital

Separable Differential Equation including Slope Fields

2017

Graphical Analysis of f' and application of FTC

Modeling using Separable Differential Equations

Particle Motion

Function Analysis from a table of data and graphically

We're not done quite yet. Here's a final reference for things we may not have covered above.
Q: I made a mistake, should I erase it?
A: If it's minor, sure. With more significant mistakes, however, it's quicker and clearer to cross out your work. You can always point the reader forward with an arrow.
Q: What if I found two solutions and I'm not sure which is correct?
A: Ideally, you'll have a hunch one way or the other and you'll clearly identify that answer. If you're flummoxed, though, or you run out of time, here's what's likely to happen. Readers will usually score both solutions, take the arithmetic mean, and then round down. You won't earn your top score, but it's possible you may earn more points. As we've seen though, with good justification, you can overcome an errant final answer.
Q: Do I have to round my answers to 3 decimal places?
A: Unless the question specifies specific rounding (usually this is to the nearest integer), you won't be penalized if you leave more than 3 correct places.
Q: Can use the nDeriv notation my calculator uses?
A: No. Stick to proper mathematics notation: f'(x) or ddx

A: Answers always require justification to receive full marks. Just justify it.
Q: What happens if I make an error in part (b) and need to use my answer in part (c)?
A: Here, the reader is on your side. You shouldn't be penalized for carrying an incorrect value from a prior part of the question. Your reader will score your answer differently. Usually, she will re-compute the sub-question with the erroneous input. If your process and justification are correct, you'll earn the corresponding points.

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AP Calc - AP Calculus Free Response Question (FRQ) Overview | Fiveable

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Study with friends using the #1 virtual studying platform for students. Create a study room, invite your buddies, & keep each other accountable.

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Derivatives of Inverse Trigonometric Functions

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Chain Rule Derivatives, Square Root, With Trig Functions, Fractions, ln, Calculus Examples

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This calculus video tutorial explains how to find derivatives using the chain rule. This lesson contains plenty of practice problems including examples of chain rule problems with trig functions, square root & radicals, fractions, ln, product rule, and quotient rule. This video gives you a simple way to find the derivative of a function using the chain rule.

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Implicit Differentiation - Find The First & Second Derivatives

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AP Calculus AB and BC Unit 3 Review [Differentiation: Composite, Implicit, and Inverse Functions]

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There are rules we can follow to find many derivatives.

For example:

The slope of a constant value (like 3) is always 0

The slope of a line like 2x is 2, or 3x is 3 etc

and so on.

Here are useful rules to help you work out the derivatives of many functions (with examples below). Note: the little mark ' means derivative of, and f and g are functions.

"The derivative of" is also written

Examples

Example: what is the derivative of sin(x) ?

From the table above it is listed as being cos(x)

It can be written as:

Or:

sin(x)' = cos(x)

Power Rule

The question is asking "what is the derivative of x ³ ?"

We can use the Power Rule, where n=3:

(In other words the derivative of x ³ is 3x ²)

So it is simply this:

"multiply by power
then reduce power by 1"

It can also be used in cases like this:

1/x is also x⁻¹

We can use the Power Rule, where n = −1:

So we just did this:

which simplifies to −1/x²

Multiplication by constant

the derivative of cf = cf'

the derivative of 5f = 5f'

We know (from the Power Rule):

So:

Sum Rule

Example: What is the derivative of x²+x3 ?

The Sum Rule says:

the derivative of f + g = f' + g'

So we can work out each derivative separately and then add them.

Using the Power Rule:

And so:

Difference Rule

What we differentiate with respect to doesn't have to be x, it could be anything. In this case v:

The Difference Rule says

the derivative of f − g = f' − g'

So we can work out each derivative separately and then subtract them.

Using the Power Rule:

And so:

Sum, Difference, Constant Multiplication And Power Rules

Using the Power Rule:

And so:

Product Rule

Example: What is the derivative of cos(x)sin(x) ?

The Product Rule says:

the derivative of fg = f g' + f' g

In our case:

We know (from the table above):

So:

the derivative of cos(x)sin(x) = cos(x)cos(x) − sin(x)sin(x)

=

Quotient Rule

To help you remember:

The derivative of "High over Low" is:
"Low dHigh minus High dLow, over the line and square the Low"
Example: What is the derivative of cos(x)/x ?

In our case:

We know (from the table above):

So:

Reciprocal Rule

The Reciprocal Rule says:
With f(x)= x, we know that f'(x) = 1
So:

Which is the same result we got above using the Power Rule.

Chain Rule

The Chain Rule says:

the derivative of f(g(x)) = f'(g(x))g'(x)

The individual derivatives are:

So:

Let's do the previous example again using that formula:

Let u = x ², so y = sin(u):

Differentiate each:

Substitute back u = x ² and simplify:

Same result as before (thank goodness!)

Another couple of examples of the Chain Rule:

The Chain Rule says:

the derivative of f(g(x)) = f'(g(x))g'(x)

The individual derivatives are:

So:
= sin(x)cos²(x)
Note: is also or many other forms.

The Chain Rule says:

the derivative of f(g(x)) = f'(g(x))g'(x)

The individual derivatives are:

So:

6800, 6801, 6802, 6803, 6804, 6805, 6806, 6807, 6808, 6809, 6810, 6811, 6812

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Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.

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Math explained in easy language, plus puzzles, games, quizzes, videos and worksheets. For K-12 kids, teachers and parents.

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Set students up for success in Calculus and beyond! Explore the entire Calculus curriculum: polynomials, derivatives, and more. Try it free!

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www.khanacademy.com Derivative of Inverse Trig Functions

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www.khanacademy.com Derivative of Inverse Trig Functions

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Implicit Differentiation - Basic Idea and Examples

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In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. With the chain rule in hand we will be able to differentiate a much wider variety of functions. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule!

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Paul's Online Notes for Inverse Trig Functions

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Paul's Online Notes for Inverse Trig Functions

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Calculus I - Derivatives of Inverse Trig Functions

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In this section we give the derivatives of all six inverse trig functions. We show the derivation of the formulas for inverse sine, inverse cosine and inverse tangent.

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Paul's Online Notes: Implicit Differentiation

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Paul's Online Notes: Implicit Differentiation

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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.

Example 1 Find \(y'\) for \(xy = 1\).

Show Solution

There are actually two solution methods for this problem.

This is the simple way of doing the problem. Just solve for \(y\) to get the function in the form that we're used to dealing with and then differentiate.

\[y = \frac{1}{x}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}y' = - \frac{1}{{{x^2}}}\]

So, that's easy enough to do. However, there are some functions for which this can't be done. That's where the second solution technique comes into play.

In this case we're going to leave the function in the form that we were given and work with it in that form. However, let's recall from the first part of this solution that if we could solve for \(y\) then we will get \(y\) as a function of \(x\). In other words, if we could solve for \(y\) (as we could in this case but won't always be able to do) we get \(y = y\left( x \right)\). Let's rewrite the equation to note this.

\[xy = x\,y\left( x \right) = 1\]

Be careful here and note that when we write \(y\left( x \right)\) we don't mean \(y\) times \(x\). What we are noting here is that \(y\) is some (probably unknown) function of \(x\). This is important to recall when doing this solution technique.

The next step in this solution is to differentiate both sides with respect to \(x\) as follows,

\[\frac{d}{{dx}}\left( {x\,y\left( x \right)} \right) = \frac{d}{{dx}}\left( 1 \right)\]

The right side is easy. It's just the derivative of a constant. The left side is also easy, but we've got to recognize that we've actually got a product here, the \(x\) and the \(y\left( x \right)\). So, to do the derivative of the left side we'll need to do the product rule. Doing this gives,

\[\left( 1 \right)y\left( x \right) + x\frac{d}{{dx}}\left( {y\left( x \right)} \right) = 0\]

Now, recall that we have the following notational way of writing the derivative.

\[\frac{d}{{dx}}\left( {y\left( x \right)} \right) = \frac{{dy}}{{dx}} = y'\]

Using this we get the following,

\[y + xy' = 0\]

Note that we dropped the \(\left( x \right)\) on the \(y\) as it was only there to remind us that the \(y\) was a function of \(x\) and now that we've taken the derivative it's no longer really needed. We just wanted it in the equation to recognize the product rule when we took the derivative.

So, let's now recall just what were we after. We were after the derivative, \(y'\), and notice that there is now a \(y'\) in the equation. So, to get the derivative all that we need to do is solve the equation for \(y'\).

\[y' = - \frac{y}{x}\]

There it is. Using the second solution technique this is our answer. This is not what we got from the first solution however. Or at least it doesn't look like the same derivative that we got from the first solution. Recall however, that we really do know what \(y\) is in terms of \(x\) and if we plug that in we will get,

\[{y}'=-\frac{{}^{1}/{}_{x}}{x}=-\frac{1}{{{x}^{2}}}\]

which is what we got from the first solution. Regardless of the solution technique used we should get the same derivative.

Example 2 Differentiate each of the following.

1. \({\left( {5{x^3} - 7x + 1} \right)^5}\), \({\left[ {f\left( x \right)} \right]^5}\), \({\left[ {y\left( x \right)} \right]^5}\)

2. \(\sin \left( {3 - 6x} \right)\), \(\sin \left( {y\left( x \right)} \right)\)

3. \({{\bf{e}}^{{x^2} - 9x}}\), \({{\bf{e}}^{y\left( x \right)}}\)

Show All Solutions Hide All Solutions
These are written a little differently from what we're used to seeing here. This is because we want to match up these problems with what we'll be doing in this section. Also, each of these parts has several functions to differentiate starting with a specific function followed by a general function. This again, is to help us with some specific parts of the implicit differentiation process that we'll be doing.

a \({\left( {5{x^3} - 7x + 1} \right)^5}\), \({\left[ {f\left( x \right)} \right]^5}\), \({\left[ {y\left( x \right)} \right]^5}\) Show Solution

With the first function here we're being asked to do the following,

\[\frac{d}{{dx}}\left[ {{{\left( {5{x^3} - 7x + 1} \right)}^5}} \right] = 5{\left( {5{x^3} - 7x + 1} \right)^4}\left( {15{x^2} - 7} \right)\]

and this is just the chain rule. We differentiated the outside function (the exponent of 5) and then multiplied that by the derivative of the inside function (the stuff inside the parenthesis).

For the second function we're going to do basically the same thing. We're going to need to use the chain rule. The outside function is still the exponent of 5 while the inside function this time is simply \(f\left( x \right)\). We don't have a specific function here, but that doesn't mean that we can't at least write down the chain rule for this function. Here is the derivative for this function,

\[\frac{d}{{dx}}{\left[ {f\left( x \right)} \right]^5} = 5{\left[ {f\left( x \right)} \right]^4}f'\left( x \right)\]

We don't actually know what \(f\left( x \right)\) is so when we do the derivative of the inside function all we can do is write down notation for the derivative, i.e. \(f'\left( x \right)\).

With the final function here we simply replaced the \(f\) in the second function with a \(y\) since most of our work in this section will involve \(y\)'s instead of \(f\)'s. Outside of that this function is identical to the second. So, the derivative is,

\[\frac{d}{{dx}}{\left[ {y\left( x \right)} \right]^5} = 5{\left[ {y\left( x \right)} \right]^4}y'\left( x \right)\]

b \(\sin \left( {3 - 6x} \right)\), \(\sin \left( {y\left( x \right)} \right)\) Show Solution

The first function to differentiate here is just a quick chain rule problem again so here is it's derivative,

\[\frac{d}{{dx}}\left[ {\sin \left( {3 - 6x} \right)} \right] = - 6\cos \left( {3 - 6x} \right)\]

For the second function we didn't bother this time with using \(f\left( x \right)\) and just jumped straight to \(y\left( x \right)\) for the general version. This is still just a general version of what we did for the first function. The outside function is still the sine and the inside is given by \(y\left( x \right)\) and while we don't have a formula for \(y\left( x \right)\) and so we can't actually take its derivative we do have a notation for its derivative. Here is the derivative for this function,

\[\frac{d}{{dx}}\left[ {\sin \left( {y\left( x \right)} \right)} \right] = y'\left( x \right)\cos \left( {y\left( x \right)} \right)\]

c \({{\bf{e}}^{{x^2} - 9x}}\), \({{\bf{e}}^{y\left( x \right)}}\) Show Solution

In this part we'll just give the answers for each and leave out the explanation that we had in the first two parts.

\[\frac{d}{{dx}}\left( {{{\bf{e}}^{{x^2} - 9x}}} \right) = \left( {2x - 9} \right){{\bf{e}}^{{x^2} - 9x}}\hspace{0.75in}\frac{d}{{dx}}\left( {{{\bf{e}}^{y\left( x \right)}}} \right) = y'\left( x \right){{\bf{e}}^{y\left( x \right)}}\]

Example 3 Find \(y'\) for the following function. \[{x^2} + {y^2} = 9\]

Show Solution

Now, this is just a circle and we can solve for \(y\) which would give,

\[y = \pm \sqrt {9 - {x^2}} \]

Prior to starting this problem, we stated that we had to do implicit differentiation here because we couldn't just solve for \(y\) and yet that's what we just did. So, why can't we use "normal" differentiation here? The problem is the "\( \pm \)". With this in the "solution" for \(y\) we see that \(y\) is in fact two different functions. Which should we use? Should we use both? We only want a single function for the derivative and at best we have two functions here.

So, in this example we really are going to need to do implicit differentiation so we can avoid this. In this example we'll do the same thing we did in the first example and remind ourselves that \(y\) is really a function of \(x\) and write \(y\) as \(y\left( x \right)\). Once we've done this all we need to do is differentiate each term with respect to \(x\).

\[\frac{d}{{dx}}\left( {{x^2} + {{\left[ {y\left( x \right)} \right]}^2}} \right) = \frac{d}{{dx}}\left( 9 \right)\]

As with the first example the right side is easy. The left side is also pretty easy since all we need to do is take the derivative of each term and note that the second term will be similar the part (a) of the second example. All we need to do for the second term is use the chain rule.

After taking the derivative we have,

\[2x + 2{\left[ {y\left( x \right)} \right]^1}y'\left( x \right) = 0\]

At this point we can drop the \(\left( x \right)\) part as it was only in the problem to help with the differentiation process. The final step is to simply solve the resulting equation for \(y'\).

\[\begin{align*}2x + 2yy' & = 0\\ y' & = - \frac{x}{y}\end{align*}\]

Unlike the first example we can't just plug in for \(y\) since we wouldn't know which of the two functions to use. Most answers from implicit differentiation will involve both \(x\) and \(y\) so don't get excited about that when it happens.

Example 4 Find the equation of the tangent line to \[{x^2} + {y^2} = 9\]

at the point \(\left( {2,\,\,\sqrt 5 } \right)\).

Show Solution

First note that unlike all the other tangent line problems we've done in previous sections we need to be given both the \(x\) and the \(y\) values of the point. Notice as well that this point does lie on the graph of the circle (you can check by plugging the points into the equation) and so it's okay to talk about the tangent line at this point.

Recall that to write down the tangent line all we need is the slope of the tangent line and this is nothing more than the derivative evaluated at the given point. We've got the derivative from the previous example so all we need to do is plug in the given point.

\[m = {\left. {y'} \right|_{x = 2,\,y = \sqrt 5 }} = - \frac{2}{{\sqrt 5 }}\]

The tangent line is then.

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

Example 5 Find \(y'\) for each of the following.

1. \({x^3}{y^5} + 3x = 8{y^3} + 1\)

2. \({x^2}\tan \left( y \right) + {y^{10}}\sec \left( x \right) = 2x\)

3. \({{\bf{e}}^{2x + 3y}} = {x^2} - \ln \left( {x{y^3}} \right)\)

Show All Solutions Hide All Solutions
a \({x^3}{y^5} + 3x = 8{y^3} + 1\) Show Solution

First differentiate both sides with respect to \(x\) and remember that each \(y\) is really \(y\left( x \right)\) we just aren't going to write it that way anymore. This means that the first term on the left will be a product rule.

We differentiated these kinds of functions involving \(y\)'s to a power with the chain rule in the Example 2 above. Also, recall the discussion prior to the start of this problem. When doing this kind of chain rule problem all that we need to do is differentiate the \(y\)'s as normal and then add on a \(y'\), which is nothing more than the derivative of the "inside function".

Here is the differentiation of each side for this function.

\[3{x^2}{y^5} + 5{x^3}{y^4}y' + 3 = 24{y^2}y'\]

Now all that we need to do is solve for the derivative, \(y'\). This is just basic solving algebra that you are capable of doing. The main problem is that it's liable to be messier than what you're used to doing. All we need to do is get all the terms with \(y'\) in them on one side and all the terms without \(y'\) in them on the other. Then factor \(y'\) out of all the terms containing it and divide both sides by the "coefficient" of the \(y'\). Here is the solving work for this one,

\[\begin{align*}3{x^2}{y^5} + 3 & = 24{y^2}y' - 5{x^3}{y^4}y'\\ 3{x^2}{y^5} + 3 & = \left( {24{y^2} - 5{x^3}{y^4}} \right)y'\\ y' & = \frac{{3{x^2}{y^5} + 3}}{{24{y^2} - 5{x^3}{y^4}}}\end{align*}\]

The algebra in these problems can be quite messy so be careful with that.

b \({x^2}\tan \left( y \right) + {y^{10}}\sec \left( x \right) = 2x\) Show Solution

We've got two product rules to deal with this time. Here is the derivative of this function.

\[2x\tan \left( y \right) + {x^2}{\sec ^2}\left( y \right)y' + 10{y^9}y'\sec \left( x \right) + {y^{10}}\sec \left( x \right)\tan \left( x \right) = 2\]

Notice the derivative tacked onto the secant! Again, this is just a chain rule problem similar to the second part of Example 2 above.

Now, solve for the derivative.

\[\begin{align*}\left( {{x^2}{{\sec }^2}\left( y \right) + 10{y^9}\sec \left( x \right)} \right)y' & = 2 - {y^{10}}\sec \left( x \right)\tan \left( x \right) - 2x\tan \left( y \right)\\ y' & = \frac{{2 - {y^{10}}\sec \left( x \right)\tan \left( x \right) - 2x\tan \left( y \right)}}{{{x^2}{{\sec }^2}\left( y \right) + 10{y^9}\sec \left( x \right)}}\end{align*}\]

c \({{\bf{e}}^{2x + 3y}} = {x^2} - \ln \left( {x{y^3}} \right)\) Show Solution

We're going to need to be careful with this problem. We've got a couple chain rules that we're going to need to deal with here that are a little different from those that we've dealt with prior to this problem.

In both the exponential and the logarithm we've got a "standard" chain rule in that there is something other than just an \(x\) or \(y\) inside the exponential and logarithm. So, this means we'll do the chain rule as usual here and then when we do the derivative of the inside function for each term we'll have to deal with differentiating \(y\)'s.

Here is the derivative of this equation.

\[{{\bf{e}}^{2x + 3y}}\left( {2 + 3y'} \right) = 2x - \frac{{{y^3} + 3x{y^2}y'}}{{x{y^3}}}\]

In both of the chain rules note that the \(y'\) didn't get tacked on until we actually differentiated the \(y\)'s in that term.

Now we need to solve for the derivative and this is liable to be somewhat messy. In order to get the \(y'\) on one side we'll need to multiply the exponential through the parenthesis and break up the quotient.

\[\begin{align*}2{{\bf{e}}^{2x + 3y}} + 3y'{{\bf{e}}^{2x + 3y}} & = 2x - \frac{{{y^3}}}{{x{y^3}}} - \frac{{3x{y^2}y'}}{{x{y^3}}}\\ 2{{\bf{e}}^{2x + 3y}} + 3y'{{\bf{e}}^{2x + 3y}} & = 2x - \frac{1}{x} - \frac{{3y'}}{y}\\ \left( {3{{\bf{e}}^{2x + 3y}} + 3{y^{ - 1}}} \right)y' & = 2x - {x^{ - 1}} - 2{{\bf{e}}^{2x + 3y}}\\ y' & = \frac{{2x - {x^{ - 1}} - 2{{\bf{e}}^{2x + 3y}}}}{{3{{\bf{e}}^{2x + 3y}} + 3{y^{ - 1}}}}\end{align*}\]

Note that to make the derivative at least look a little nicer we converted all the fractions to negative exponents.

Example 6 Assume that \(x = x\left( t \right)\) and \(y = y\left( t \right)\) and differentiate the following equation with respect to \(t\). \[{x^3}{y^6} + {{\bf{e}}^{1 - x}} - \cos \left( {5y} \right) = {y^2}\]

Show Solution

So, just differentiate as normal and add on an appropriate derivative at each step. Note as well that the first term will be a product rule since both \(x\) and \(y\) are functions of \(t\).

\[3{x^2}x'{y^6} + 6{x^3}{y^5}y' - x'{{\bf{e}}^{1 - x}} + 5y'\sin \left( {5y} \right) = 2yy'\]

There really isn't all that much to this problem. Since there are two derivatives in the problem we won't be bothering to solve for one of them. When we do this kind of problem in the next section the problem will imply which one we need to solve for.

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In this section we will discuss implicit differentiation. Not every function can be explicitly written in terms of the independent variable, e.g. y = f(x) and yet we will still need to know what f'(x) is. Implicit differentiation will allow us to find the derivative in these cases.