# data[].attributes.body
I don't know how to use the picture to solve the problem. Also, will you take down this site once the next semester comes around?

P.S. I will definitely miss this class, it was a lot of fun doing math. The semester went by really fast since I've grown fond of the class. It's unfortunate that all good things must come to an end but why so quickly?

# data[].attributes.body
I don't know how to solve the problem.

# data[].attributes.body
I don't know how to compute the lower Riemann sums for a trig. function.

# data[].attributes.body
How do I solve this type of problem?

# data[].attributes.body
I'm not sure how to solve this problem. Also, what does \(n = 6\) mean?

# data[].attributes.body
I hope what we did in class today clarified your doubts. If not, ask again. I will try to check this website every day.

# data[].attributes.body
I don't know how to solve this problem so can show the steps in solving this problem?

# data[].attributes.body
How do I solve this problem?

# data[].attributes.body
I'm not sure how to find A, B, C so can you show the steps to solving problems like these?

# data[].attributes.body
I don't know how to solve this problem completely.

Also, how should I use y'' - 4y = 0 in solving A, B and C?

# data[].attributes.body
remember that arcsin is the inverse of the sin function on the domain (-pi/2,pi/2)

# data[].attributes.body
I know how to find the derivative of this, it should be \(3e^{arcsin(3x)}/\sqrt{1-9x^2}\) but once I plug in the 1/6 I can't figure out how to simplify it correctly.

# data[].attributes.body
Remember that \(\ln 1=0\)

# data[].attributes.body
Do the concepts (1/0)=infinity and (1/infinity)=0 only apply to limits as x approaches to 0 or infinity, or does it apply to all limits? For example, in the limit below, would the answer be DNE or infinity since you get (1/0)?

Lim as x--> 1 of (1-lnx)/(x-1)

# data[].attributes.headline
Use Differentials to Estimate the Value of the Expression

# data[].attributes.subject
Use Differentials to Estimate the Value of the Expression

# data[].attributes.body
Hello there Adrian, I was just a bit confused on what x-value I'm supposed to use in this differential problem. I'm tempted to use 64 since it's a perfect cube root (no other number is a perfect cube root close to 84) but isn't that too far of a number?

Expression: \(84^(1/3)\)

# data[].attributes.body
How do I solve this problem?

# data[].attributes.body
Use the definition of cosh in terms of the exponential function and remember that the exponential function is the inverse of the natural logarithm. Take the derivative after performing the transformations and simplifications above.

# data[].attributes.body
How do I differentiate this equation?

# data[].attributes.body
Compute \(y'\) and \(y''\) in terms of \(A,B,C,x\).
Then write down the three equations that the question imposes as constraint.
Derive from this what A, B and C have to be, remembering that C>0.

# data[].attributes.body
I don't know where to start in solving this problem

# data[].attributes.headline
Answer on rectangle inscribed inside semi-circle

# data[].attributes.subject
Answer on rectangle inscribed inside semi-circle

# data[].attributes.body
Ok, I see that this question is hard. I will give you hints, but not the full answer.
1. Make sure you draw on a sheet of paper a semi-circle (It is easier to draw the top-half of a circle for what comes next), and a rectangle inscribed in it.

Do you understand what inscribed means? Did you notice that your rectangle has to be symmetric about the vertical line (picture it in your head, or even better, draw it as well) that cuts the semicircle in half?

2. Give names to the height and length of your rectangle. I pick a and b.
3. Define an area variable, A. Can you express A as a function of a and b?
4. Now, do you notice that your rectangle can be "controlled" by another variable yet? It is an angle variable. It is the angle formed by the upper-right corner of your rectangle, with the horizontal baseline of the semicircle. Call that angle x for instance.

Can you figure out the allowable domain for x?

5. Your task now is to express a as a function of x, and b as a function of x. Hint: these functions are trig functions...but be careful, your circle is not a unit circle, so there is an additional mutliplicative constant somewhere.
6. If you managed to do all of the above, then you should be able to write A as a function of x only. Then it is just a matter of differentiating A with respect to x...
... Good luck. Think hard.

# data[].attributes.headline
A link for advanced discussion on the power of a number

# data[].attributes.subject
A link for advanced discussion on the power of a number

# data[].attributes.body
http://mathforum.org/library/drmath/view/76619.html

# data[].attributes.body
I know that this question was asked before, but I just don't know how to use the radius in the semicircle to find the area of the rectangle.

# data[].attributes.body
What are the main dimension variables of a cylinder? The dimensions that characterize it uniquely? Well, there is the radius of its base, and its height right? Lets call them \(r\) and \(h\). For each value of these two variables, you will get a different cylinder, which will presumably have a different cost. So one way to start is to set up a cost function of 2 variables for the total cost of your cylinder: \(C(r,h)\). But remember that you know the total capacity of the cylinder. So use the formula for the volume of the cylinder to express \(h\) in term of \(r\). This way, \(C(r,h)\) may be reduced to a function of \(r\) only: \(C(r)\).

Next, you are given information about how the cost should be subdivided. The cost of the whole cylinder will be a sum of the costs of three subparts: bottom base, side, top base (do you picture those parts?). Presumably, the area of the bottom and top parts have nothing to do with the height of the cylinder, so their respective cost functions, call them \(C_{bot}(r)\) and \(C_{top}(r)\) are only functions of \(r\).

I will write \(C_{bot}(r)\) and you deduce the other one on your own.
\(C_{bot}(r)=20\cdot \pi r^2\).

Now, concerning the cost of the side, you will need both \(r\) and \(h\).
As for the bottom, find out the formula for the area of the side in terms of these two variables and multiply it by the corresponding cost for this part of the cylinder. And from what I said above, you should be able to replace \(h\) in the formula by an expression involving \(r\).

So, at this point, you should have \(C(r)=C_{bot}(r)+C_{top}(r)+C_{side}(r)\), and you should have explicit expressions i \(r\) for all terms.

Your final task is to find the local minimum of \(C\) and give the \(r\) and \(h\) values that correspond to this cost.

Does this make sense?

# data[].attributes.body
Remember that \(a^2-b^2=(a+b)(a-b)\) and that for any expression the   following holds:\(u (x)=\sqrt{u (x)}\sqrt{u (x)}\). Also try to compute \(11^2\).

# data[].attributes.body
Hello there Adrian, could you detail the process of how the equation on the left simplifies to the equation on the right in the image? Thanks.

# data[].attributes.body
I'm not sure how to include the costs in finding the dimensions of the cylinder

# data[].attributes.body
1. Make a sketch of what this mean. Could you?
2. Label the important variables in your sketch and identify the ones that are allowed to vary in the problem, and the ones that are fixed.
3. One of these variables has to be the area of the rectangle. Let's call it A. A will be a function of some other variables. You want to find the absolute maximum of A. But first find its domain.

Does all this help at all? Really try to do the steps I explained above in that order.

# data[].attributes.body
how do I solve this?

# data[].attributes.body
Section 4.1, Problem 14
The derivative is \(f'(x)=3x^2-12x+12\) and is defined on \(\mathbb{R}\). Its sign is the same as the sign of \(x^2-4x+4=(x-2)^2\), which is positive on \(\mathbb{R}\setminus \{2\}\) and vanishes at \(x=0\). This tells us that \(f\) is strictly increasing on \(\mathbb{R}\) and therefore has an inverse function.

Section 4.1, Problem 20
The function \(f\) is undefined at \(x=0\). For any \(a,b\neq 0\), we have,\begin{align}
f(a)=f(b)&\Rightarrow\frac{a+2}{a}=\frac{b+2}{b}\\
&\Rightarrow a(b+2)=b(a+2)\\
&\Rightarrow ab+2a=ba+2b\\
&\Rightarrow 2a=2b\\
&\Rightarrow a=b.
\end{align}
This tells us that \(f\) is one-to-one.
We can already tell that the domain of \(f^{-1}\) is the range of \(f\), namely \(\mathbb{R}\setminus \{1\}\) (try to graph \(f\); analyzing its asymptotes and derivative will help).

As for the inverse of \(f\), we derive its formula as follows:\begin{align}
x=\frac{y+2}{y} &\Rightarrow xy=y+2\\
&\Rightarrow xy-y=2\\
&\Rightarrow y(x-1)=2\\
&\Rightarrow y=\frac{2}{x-1}.
\end{align}

Section 4.2, Problem 16
We want to find the derivative of \(f(x)=e^{x^2\sin x}\). We use logarithmic differentiation. For this we take the natural log of both sides of the equation of \(f\) which has the effect of annihilating the exponential: \(\ln f(x)=x^2\sin x\). Next, we differentiate both sides with respect to \(x\):
\begin{equation}
\frac{f'(x)}{f(x)}=\frac{d}{dx}\left(x^2\sin x\right)=2x\sin x + x^2\cos x.
\end{equation}
Rearranging yields the result:
\begin{equation}
f'(x)=f(x)\left(2x\sin x + x^2\cos x\right)=\left(2x\sin x + x^2\cos x\right)e^{x^2\sin x}.
\end{equation}

Section 4.2, Problem 20
We want to find the derivative of \(f(x)=\sin\left(e^{5x-1}\right)\). We need the chain rule, and the fact that \(\left(e^u\right)'=u'e^u\).
We get, \begin{equation} f'(x)=5e^{5x-1}\cos\left(e^{5x-1}\right).\end{equation}

Section 4.2, Problem 30
We want to find the slope of the tangent line to the graph of \(f\) at \(c=0\), where \begin{equation}f(x)=\frac{e^{2x}}{1+e^x}\end{equation}.
We use the quotient rule to find the derivative of \(f\):
\begin{align}
f'(x)=\frac{2e^{2x}(1+e^x)-e^{2x}e^x}{\left(1+e^x\right)^2}.
\end{align}
Substituting in \(0\) for \(x\) yields,
\(f'(0)=\frac{2(1+1)-1\cdot 1}{(1+1)^2}=3/4\).

# data[].attributes.body
Rewrite: \(f(x)=e^{2x\ln x}\), so that \(f(x)=e^u\). Now recall that \(f'(x)=u'e^u\). We first compute \(u'\). By the product rule we get:
\(u'=2\ln x +\frac{2x}{x}=2(\ln x +1)\). Hence, since \(e^u=f(x)=x^{2x}\) we find,
\begin{equation}
f'(x)=2(\ln x+1)x^{2x}.
\end{equation}
Evaluating the latter at \(x=1\) yields \(f'(1)=2\).

# data[].attributes.headline
Question on an EMCF that I need help solving 

# data[].attributes.subject
Question on an EMCF that I need help solving 

# data[].attributes.headline
How to solve for \(y\) in \(y^2=39^2-36^2\) without a calculator

# data[].attributes.subject
How to solve for \(y\) in \(y^2=39^2-36^2\) without a calculator

# data[].attributes.body
Recall my favorite formula: \(a^2-b^2=(a+b)(a-b)\). Hence, \(y^2=(39+36)(39-36)=3\cdot(13+12)\cdot 3=9\cdot 25\). So \(y=\pm 3\cdot 5=\pm 15\).

# data[].attributes.body
Let's start by deriving g, we get, \(g'=-\frac{\left(f^{-1}\right)'}{\left(f^{-1}\right)^2}\). Plugging in 6 yields:\(g'(6)=-\frac{\left(f^{-1}\right)'(6)}{\left(f^{-1}(6)\right)^2}\). The denominator is \((-7)^2=49\) and the numerator is \(3/4\). So the final answer is \(g'(6)=\frac{3}{4\cdot 49}\)

# data[].attributes.body
Hi Adrian,

this was a question from an online quiz and I just don't get what to do or where to start.

# data[].attributes.body
https://dl.dropboxusercontent.com/u/90951391/key_whw7_AdrianRadillo.pdf

# data[].attributes.body
I will try to answer your questions by Friday

# data[].attributes.body
I assume that the pythagorean theorem should be used to relate the variables, but would the rates of change for the cars be negative or positive? if you could solve this it would be much appreciated!

# data[].attributes.body
I don't think we did a problem like this before so how should I solve this problem?

# data[].attributes.body
You are right about the formula for the first derivative, but since the question is about concavity, you should compute the second derivative: \(f''(x)=-2(1+2\cos(2x))\), which may be rearranged into, \[f''(x)=-8(\cos x +1/2)(\cos x -1/2).\] Can you analyze the sign of \(f''\) from there and answer the question?

# data[].attributes.headline
For each function, find
the interval(s) on which the graph is concave up (if any), find the interval(s) on which
the graph is concave down (if any), find the points of inflection (if any).

# data[].attributes.subject
For each function, find
the interval(s) on which the graph is concave up (if any), find the interval(s) on which
the graph is concave down (if any), find the points of inflection (if any).

# data[].attributes.body
\(f(x)=2cos^2(x)-x^2\)
I've found the derivative to be \(f'(x)=-2(x+sin2x)\), but am not sure how to proceed when I set it equal to zero.

# data[].attributes.body
Make sure you know your definitions!

# data[].attributes.body
QUESTION: Find any extreme values for \(f(x)=x/(1+x^2)\) and label them as local maxima or minima.

ANSWER: We first compute the derivative of \(f\) (using the quotient rule), which is defined everywhere since the denominator of \(f\) never vanishes:
\begin{equation}
f'(x)=\frac{1+x^2-2x^2}{(1+x^2)^2}=\frac{1-x^2}{(1+x^2)^2}.
\end{equation}
Remember that for extreme values, the signs of the first derivative is what matters. Given the formula for \(f'\) above, we notice that \(f'(x)\) has always the same sign as its numerator (because the denominator is always positive). The numerator of \(f'\) is a quadratic which is zero at \(-1\) and \(1\), positive on \(-1,1\) and negative otherwise. Hence, the derivative of \(f\) changes sign at \(-1\) from negative to positive, and at \(1\) from positive to negative.
The first derivative test tells us that \(f(-1)=-1/2\) is a local minimum and \(f(1)=1/2\) is a local maximum.

# data[].attributes.body
Differentiate the equation for the area, with respect to \(r\), on both sides to get:
\begin{equation}
\frac{dA}{dr}=\frac{d}{dr}\left( \frac{1}{2}r^2\theta\right).
\end{equation}
Since the area is constant, its derivative is zero. Using the product rule on the right-hand side we get:
\begin{equation}
0=\frac{1}{2}\left(2r\theta + r^2\frac{d\theta}{dr}\right).
\end{equation}
Rearranging to get the change rate for \(\theta\) on one side we get,
\begin{equation}
\frac{d\theta}{dr}=\frac{-2\theta}{r}.
\end{equation}
Now, we notice that \(\theta\) itself may be expressed in terms of \(A\) when \(r=5\), using the equation given in the question. That is, we find,
\(\theta=2A/25\). Using this fact, together with \(r=5\), we find,
\begin{equation}
\frac{d\theta}{dr}=-\frac{2A}{125}.
\end{equation}
We can't go much further, unless you know how to solve first order linear differential equations...
Let me know if this does not make sense.

# data[].attributes.headline
I'm reviewing for the test by doing some odd problems from the textbook, could you explain how to solve this related rates problem?

# data[].attributes.subject
I'm reviewing for the test by doing some odd problems from the textbook, could you explain how to solve this related rates problem?

# data[].attributes.body
The free fall equation, with \(v_0\) given, is:
\begin{equation}
h(t)=-16t^2+4t+h_0.
\end{equation}
We are told that at a specific time \(t^*\) the height is \(h(t^*)=90\) and 2 seconds later, the height is \(h(t^*+2)=10\). You are asked to find \(h_0\).
Here is how I do it. First find \(t^*\). In order to do this, we use the following equation (do you understand why it is true?)
\begin{equation}
\frac{h(t^*+2)-h(t^*)}{2}=\frac{10-90}{2}=-40.
\end{equation}
The numerator of the left-hand side simplifies to
\begin{equation}
h(t^*+2)-h(t^*)=-16(4+4t^*-1/2).
\end{equation}
So we want to solve the following equation for \(t^*\):
\begin{equation}
\frac{-16(4+4t^*-1/2)}{2}=-40.
\end{equation}
The solution is \(t^*=3/8\) seconds.

The next step is to find \(h_0\) using the height equation:
\begin{equation}
h(t^*)=90=-16(t^*)^2+4t^*+h_0.
\end{equation}
Solving for \(h_0\) yields \(h_0=363/4\) feet.

Does it make sense?

# data[].attributes.body
How would I solve this problem?

# data[].attributes.body
I spoke with the instructors and the unfortunate expression "canonical cup" will be changed to "conical cup" in the textbook, as we discussed in class. Also, the rate at which water is leaking should be specified in volume units per seconds, rather than in height units per seconds.

# data[].attributes.body
Hey Adrian,
In our written homework, the problem specifies "Water is leaking from the bottom of a canonical cup that is 6 inches across and 8 inches deep." What in the world is a canonical cup? Is it a cone? A cylinder? Something more complex?

# data[].attributes.body
The question was, "How many tangent lines to the curve \(y=x/(x+1)\) pass through the point \((1,2)\)? At which points do these tangent lines touch the curve?"

This question is designed to test two things: 1. your ability to think about the geometry of the graph of a function and 2. your ability to translate a geometrical problem into an algebraic one, i.e., an equation to solve.

Here is one approach to the solution. The first thing that I strongly encourage you to do is to try to sketch the curve, without any computer or calculator. Here are a few hints about the graph:
1. The limit of \(y\) as \(x\) tends to both \(+\infty\) and \(-\infty\) is \(1\), therefore the graph has a horizontal asymptote at both ends of the \(x\)-axis.
2. The function \(f(x)=x/(x+1)\) is a rational function with denominator that vanishes at \(-1\), the graph will therefore have a vertical asymptote at \(x=-1\).
3. Then, analyze the sign of \(y\) for values of \(x\) close to \(-1\). When \(x<-1\), both numerator and denominator of \(f(x)\) are negative, so \(f(x)\) is positive and this means that \(\lim_{x\to -1^-}f(x)=+\infty\). On the other hand, for values of \(x\) close to \(-1\) but satisfying \(x>-1\), the denominator will become positive while the numerator will remain negative, hence \(\lim_{x\to -1^+}f(x)=-\infty\).

Can you sketch the curve now?

Do you understand geometrically what the problem is asking?

From your sketch, can you get an idea of the answer?

Now, here is how to solve the question formally. The derivative of \(f\)
is \(f'(x)=\frac{1}{(1+x)^2}\) (quotient rule). We will now consider the family of tangent lines to the graph of \(f\). To each point \((\alpha,\beta)\) of the graph of \(f\) corresponds a tangent to this graph (at that point). Call this tangent line \(L_\alpha\). Can you find the equation of \(L_\alpha\)?

The line \(L_\alpha\) has slope \(f'(\alpha)\) and passes through the point \(\alpha,\beta\), by construction. Therefore the line has equation:
\begin{equation}
y-\beta=f'(\alpha)(x-\alpha).
\end{equation}
Note that, by assumption, \(\alpha,\beta\) is on the graph of \(f\), which precisely means that \(\beta=f(\alpha)=\alpha/(1+\alpha)\). Therefore, using all the information listed so far, the equation of the line \(L_\alpha\) now becomes,
\begin{equation}
y-\frac{\alpha}{1+\alpha}=\frac{1}{(1+\alpha)^2}(x-\alpha).
\end{equation}
Are you with me so far? To summarize, for each point \((\alpha,\beta)\) lying on the graph of \(f\) we can express the equation of the tangent line to the graph through that point as:
\begin{equation}
y=\frac{x-\alpha}{(1+\alpha)^2}+\frac{\alpha}{1+\alpha}.
\end{equation}

Now the problem may be rephrased as follows: For how many values of \(\alpha\) does the line \(L_\alpha\) pass through the point \((1,2)\)?

Well, the point \((1,2)\) belongs to \(L_\alpha\) if and only if the coordinates of the point satisfy the equation of the line, that is, if and only if the following equation is true:
\begin{equation}
2=\frac{1-\alpha}{(1+\alpha)^2}+\frac{\alpha}{1+\alpha}.
\end{equation}

So all that is left to do is to solve the previous equation for \(\alpha\).
Recall that we must have \(\alpha \neq -1\), so we may multiply left and right-hand sides of the previous equation by \((1+\alpha)^2\) to get,
\begin{align}
2(1+\alpha)^2&=1-\alpha+\alpha(1+\alpha)\\
\Leftrightarrow 2+2\alpha^2+4\alpha&=1+\alpha^2
\Leftrightarrow \alpha^2+4\alpha+1&=0.
\end{align}
The previous quadratic has two real roots:
\begin{equation}\alpha=\frac{-4\pm\sqrt{16-4}}{2}=-2\pm \sqrt{3}.\end{equation}

So the only two points on the graph of \(f\) at which the tangent line passes through \((1,2)\) are \(\left((-2+\sqrt{3},f\left(-2+\sqrt{3}\right)\right)\) and \(\left(-2-\sqrt{3},f\left(-2-\sqrt{3}\right)\right)\).

# data[].attributes.body
The question was to find \(\frac{dy}{dx}\) given that \begin{equation}y=\sqrt{x+\sqrt{x+\sqrt{x}}}.\end{equation}
To answer, I chose to break up the problem in parts. Let \(u(x)=\sqrt{x}\), \(v(x)=x+\sqrt{x}\) and \(w(x)=(u\circ v)(x) =\sqrt{x+\sqrt{x}}\). Then, we have \(y=u(x+w(x))\) and by the chain rule \begin{equation}\frac{dy}{dx}=(1+w'(x))\cdot u'(x+w(x)).\end{equation}
Let's find the terms involved.
First, \(u'(x)=\frac{1}{2\sqrt{x}}\) and \(v'(x)=1+u'(x)\), so that by the chain rule,
\begin{equation}
w'(x)=v'(x)\cdot (u'\circ v)(x)=\left(1+\frac{1}{2\sqrt{x}} \right)\cdot \frac{1}{2 w(x)}.
\end{equation}
If you want, you may rearrange this into,
\begin{equation}w'(x)=\frac{2\sqrt{x}+1}{4\sqrt{x^2+x\sqrt{x}}},\end{equation}
or into,
\begin{equation}
w'(x)=\frac{2x^\frac{1}{2}+1}{4\left(x^2+x^\frac{3}{2}\right)^\frac{1}{2}}.
\end{equation}
We can now come back to \(dy/dx\):
\begin{align}
\frac{dy}{dx}=\left(1+\left(1+\frac{1}{2\sqrt{x}} \right)\cdot \frac{1}{2 w(x)}\right)\cdot \frac{1}{2\ y}
\end{align}

Unfortunately, the expression doesn't get much nicer than this. In math, sometimes, one has to accept that formulae are ugly and remain like this. The only way to keep the formulae readable is to rename variables to replace big expressions by single letters, as I did above with \(w(x)\) and \(y\).

# data[].attributes.body
I will solve the modified problem: \(y=2x+\frac{7}{x}\).
\(y\) is a sum of two terms, so \(dy/dx\) is the sum of the derivatives of each term. The first term has derivative \(2\) and the second term has derivative \(-\frac{7}{x^2}\), so \(\frac{dy}{dx}=2-\frac{7}{x^2}\).

# data[].attributes.body
How would I find \(\frac{dy}{dx}\) in this equation?

\(y=x-\frac{3}{x}\)

# data[].attributes.headline
Answer to still another one of the questions below

# data[].attributes.subject
Answer to still another one of the questions below

# data[].attributes.body
If \(f(x)=(1+x^2)^{-2}\), then you should find that,
\begin{equation}
f'(x)=\frac{-4x}{(1+x^2)^{3}}.
\end{equation}
Note that it is a quotient of the form \(a/b\), with \(b\) always positive. So the sign of this derivative is the same as the sign of \(-x\), that it is, it is \(0\) when \(x=0\), it is positive when \(x<0\) and it is negative when \(x>0\)

# data[].attributes.headline
Answer to another one of the questions below

# data[].attributes.subject
Answer to another one of the questions below

# data[].attributes.body
Given \(f(x)=\tan^2(x^4+1)\), we use the chain rule to find the derivative, with:
\begin{align}
&u(x)=\tan^2(x)\qquad u'(x)=2\sec^2(x)\tan(x)\\
&v(x)=x^4+1\qquad v'(x)=4x^3\\
&f(x)=(u\circ v)(x) \qquad f'(x)=v'(x)\cdot (u'\circ v)(x)\\
&f'(x)=8x^3\sec^2\left(x^4+1\right) \tan\left(x^4+1\right)
\end{align}

# data[].attributes.headline
Answer to one of the previous questions below

# data[].attributes.subject
Answer to one of the previous questions below

# data[].attributes.body
The calculations are very lengthy, you will need to derive twice with respect to \(x\). The first derivation yields,
\begin{align}
2xy^2+2yy'x^2-2&=0\\
\Rightarrow xy^2+y'yx^2&=1\\
\Rightarrow y'&=\frac{1-xy^2}{x^2y}\\
\Rightarrow y'&=\frac{1}{x^2y}-\frac{y}{x}
\end{align}
Now we derive a second time with respect to \(x\) (my calculations might contain some typos):
\begin{align}
\frac{d^2y}{dx}&=\frac{d}{dx}(y')\\
&=-\frac{2xy+x^2y'}{(x^2y)^2}-\frac{y'x-y}{x^2}\\
&=-\left[\frac{2}{x^3y}+\frac{y'}{x^2y^2}+\frac{y'}{x}-\frac{y}{x^2} \right]\\
&=-\left[\frac{2}{x^3y}+\frac{1}{x^4y^3}-\frac{y}{x^3y^2}+\frac{1}{x^3y}-\frac{y}{x^2} -\frac{y}{x^2}\right]\\
&=-\frac{2}{x^3y}-\frac{1}{x^4y^3}+\frac{2y}{x^2}
\end{align}

# data[].attributes.body
The numerator may be rewritten:
\(\frac{\cos^2(5x)-1}{\cos^2(5x)}=\frac{-\sin^2(5x)}{\cos^2(5x)}\),
so the whole expression reads:
\(-\left(\frac{\sin(5x)}{3x}\right)^2\cdot \frac{1}{\cos^2(5x)}\),
and using the known trig limits we get,
\(\lim_{x\to 0}\frac{1-\sec^2(5x)}{(3x)^2}=-\frac{25}{9}\).

# data[].attributes.body
How do I exactly solve this problem?

# data[].attributes.body
\(x^2y^2-2x=3\)

# data[].attributes.body
\(f(x)=tan^2(x^4+1)\)

# data[].attributes.headline
find the tangent line to the graph of the given equation at the indicated point 

# data[].attributes.subject
find the tangent line to the graph of the given equation at the indicated point 

# data[].attributes.body
I am not sure what to do after finding the derivative of f(x). What would be the next step?

# data[].attributes.headline
Model Answer Quiz from Friday 16 Sep 201

# data[].attributes.subject
Model Answer Quiz from Friday 16 Sep 201

# data[].attributes.body
I will solve the question for \(f(x)=\csc(4x)\) and \(x=\frac{-\pi}{12}\). First, the derivative is \(f'(x)=-4\csc(4x)\cot(4x)=-4\frac{\cos(4x)}{\sin^2(4x)}\). Next, the slope of the tangent line at the given point will be \[
f'\left(\frac{-\pi}{12}\right)=4\cdot \frac{\frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{8}{3},
\]
we deduce that the slope of the normal line, which is always perpendicular to the tangent, is \(-\frac{3}{8}\).
It now only remains to find the point \((x,f(x))\) through which both lines pass. We have,
\[
f\left(\frac{-\pi}{12}\right)=\frac{1}{\sin\left(\frac{-\pi}{3}\right)}=-\frac{2}{\sqrt{3}}
\]
The equation of the tangent line is therefore:
\[
y=\frac{8}{3}\left(x+\frac{\pi}{12}\right)-\frac{2}{\sqrt{3}}.
\]
And the equation of the normal line is:
\[
y=-\frac{3}{8}\left(x+\frac{\pi}{12}\right)-\frac{2}{\sqrt{3}}.
\]
It would be good to rearrange these equations in standard form in an exam.

# data[].attributes.headline
find the equations of the tangent and normal lines to the curve f at the given point

# data[].attributes.subject
find the equations of the tangent and normal lines to the curve f at the given point

# data[].attributes.headline
Review session for test 2 on Monday 19 Sept. in room ALC, Cougar Place 1020

# data[].attributes.subject
Review session for test 2 on Monday 19 Sept. in room ALC, Cougar Place 1020

# data[].attributes.body
Today's quiz will be on sections 2.3 and 2.4 of the textbook.

# data[].attributes.body
What will the quiz tomorrow be covering?

# data[].attributes.body
I will modify the question, as usual,
\(f(x)=\cos\left(\frac{-x}{3x-2}\right)\).
We need chain rule and quotient rule.
Let's work on this piece by piece.
The quotient rule gives,
\begin{align}
\frac{d}{dx}\left(\frac{-x}{3x-2}\right)&=\frac{-1\cdot(3x-2)+x\cdot 3}{(3x-2)^2}=\frac{2}{(3x-2)^2}
\end{align}
Now the chain rule gives,
\begin{align}
\frac{d}{dx}\left(\cos\left(\frac{-x}{3x-2}\right)\right)&=-\frac{2}{(3x-2)^2}\sin\left(\frac{-x}{3x-2}\right)
\end{align}
At this point we are done.

# data[].attributes.body
You must also know \(\cos' (x)=-\sin (x)\), \(\sin'(x)=\cos (x)\) and \(\tan'(x)=\sec^2 (x)\).

# data[].attributes.body
I will solve the question for the modified function \(f (x)=-2x^2+7\) and line \(y=6x+1\).
First compute \(f'(x)=-4x\). Then, since we want a tangent parallel to the given line, we must have \(f'(x)=6\) (because the slopes of 2 parallel lines are always equal).
We then solve for \(x\) the last equation that yields: \(x=-3/2\). The last step is to compute \(f (-3/2)=5/2\).
The final answer is that there is only one point meeting the condition:\((-3/2,5/2)\).

# data[].attributes.body
I think that first you take the derivative of f(x), which is f'(x) = 2x, but I'm not sure what the next step is

# data[].attributes.body
I'll try to upload solutions to these asap.

# data[].attributes.body
I will modify the question to the following: if \(f(x)=\sin^2(-x)\), find \(f'(\pi/6)\).
To answer, you have to take two steps:
1. First compute the derivative of \(f\). With my example, let's call \(u:=\sin(-x)\). Then \(f'=\frac{d}{dx}\left(u^2\right)\). Recall that, since \(u\) is a function of \(x\), you want to use the rule \(\frac{d}{dx}\left(u^n\right)=nu'u^{n-1}\). So here, since by the chain rule \(u'=-\cos(-x)\), we conclude that \begin{equation}f'(x)=-2\cos(-x)\sin(-x).\end{equation}

If you want, taking into account the fact that \(\cos\) is an even function and \(\sin\) is odd, you may further simplify: \(f'(x)=2\sin(x)\cos(x)=\sin(2x)\).

2. The second step is to substitute \(x\) for \(\pi/6\) to get:\(f'(\pi/6)=\sin(\pi/3)=\sqrt{3}/2\).

Ask about anything that does not make sense to you.

# data[].attributes.body
If \(f(x)=\cos^2(2x)\), find \(f'(\pi/3)\)

# data[].attributes.body
I will solve the question for modified values: \begin{equation}\lim_{x\to 0}\frac{\cos(4x)-1}{7x}.\end{equation} The term \(\cos(4x)-1\) on top can be re-written \(-(1-\cos(4x))\). Hence, \begin{align}\frac{\cos(4x)-1}{7x}&=-\frac{1-\cos(4x)}{7x}.\end{align}
Are you with me so far? Then a last transformation into, \begin{equation}-\frac{4}{7}\cdot\frac{1-\cos(4x)}{4x},\end{equation} allows us to use the known limit \(\lim_{x\to 0 }\frac{1-\cos(ax)}{ax}=0\) to deduce that, the limit of your initial expression is \(0\).

# data[].attributes.body
How do I go about solving this problem?

# data[].attributes.body
The following is very important:
If you know that a line goes through the point \((a,b)\) and has slope \(m\), are you able to write down the equation of the line?
You must learn the answer. The answer is yes and the equation is:\begin{equation}\frac{y-b}{x-a}=m.\end{equation} You may rearrange the previous equation in standard form (\(y=\alpha x+\beta\)) if you like.

Do you see how this applies to your question? The point is \((1,2)\) and the slope is \(5\).

# data[].attributes.body
If \(f(1)=2\) and \(f'(1)=5\), what is the equation of the line that is tangent to f(x) at x=1?

# data[].attributes.body
I will answer with modified values. Say \(f=-2\cos(x)\) and \(c=\pi/3\), then the first thing to do is to compute the difference quotient:
\begin{align}
\frac{f(c+h)-f(c)}{h}&=\frac{-2\left(\cos\left(\frac{\pi}{3}+h\right)-\cos\left(\frac{\pi}{3}\right)\right)}{h}\\
&=-\frac{2}{h}\left(\cos\left(\frac{\pi}{3}\right)\cos(h)-\sin\left(\frac{\pi}{3}\right)\sin(h)-\cos\left(\frac{\pi}{3}\right)\right)\\
&=-\frac{2}{h}\left(\frac{1}{2}(\cos(h)-1)-\frac{\sqrt{3}}{2}\sin(h)\right)\\
&=\frac{1-\cos(h)}{h}+\sqrt{3}\frac{\sin(h)}{h}
\end{align}
We deduce that the limit is \(\sqrt{3}\). Does this make sense to you?

# data[].attributes.body
Given \(f(x)=5cos(x)\) and \(c=7π/6\), determine whether \(\lim_{h\to 0}\frac{f(c+h)−f(c)}{h}\)exists, and give the value if the limit does exist.

# data[].attributes.body
I will solve the problem for the modified function: \(f(x)=-2x^2+9/x\).

1. Find the derivative
Here it is: \(f'(x)=-3x-\frac{9}{x^2}\).

2. Solve the equation \(f'(x)=0\) (because this is what it means to say that the tangent is horizontal):
The derivative is not defined at \(x=0\). When \(x\neq 0\) we have:
\begin{align}
&-3x-\frac{9}{x^2}=0& \\
\Rightarrow & -3x^3-9=0& \\
\Rightarrow & x^3=-3&\\
\Rightarrow & x=-3^\frac{1}{3}&
\end{align}

3. The previous equation only had 1 solution, so there will only be one horizontal tangent line to the graph of \(f\). This tangent must have an equation of the form \(y=c\), where \(c\) is a constant number. Since the tangent passes through the point \(\left(-3^\frac{1}{3},f\left(-3^\frac{1}{3}\right)\right)\), we must have \(c=f\left(-3^\frac{1}{3}\right)\). In your exercise, compute explicitly the right-hand side of this last equation.

# data[].attributes.headline
Consider the function \(f\), find the \(x\)-value(s) where the tangent line is
horizontal. Find the equation of each horizontal tangent line. 

# data[].attributes.subject
Consider the function \(f\), find the \(x\)-value(s) where the tangent line is
horizontal. Find the equation of each horizontal tangent line. 

# data[].attributes.body
\(f(x)=x^2-(16/x)\)

# data[].attributes.body
I will solve it with modified values: \(f(x)=-x^9+3x^7-x^5+2x+5\); \(c=1\). The derivative is: \(f'(x)=-9x^8+21x^6-5x^4+2\).
Slope: you first need to provide the slope of the tangent. It is nothing else than the value of the derivative evaluated at \(c\). With my example the slope is \(f'(c)=f'(1)=-9+21-5+2=9\).
Equation of tangent line: now you only have to find the equation of the tangent. There are several ways of doing it. Here is one way. The equation will always be of the form \(y=ax+b\). Since we know the slope, \(a=9\), we only have to find \(b\). To do this, we use the fact that the tangent line passes through the point \((c,f(c))\) (see the question). We hence rewrite the equation of the tangent line, but instead of \(y\) we write \(f(c)=f(1)\) (and compute it), and instead of \(x\), we write \(c=1\). Let's first quickly compute \(f(c)=f(1)=-1+3-1+2+5=8\). The equation of the tangent line evaluated at \((1,f(1))\) now reads: \(8=9\cdot 1+b\). We solve for \(b\) and get \(b=-1\). We now only have to rewrite the equation of the tangent in its general form with the values of \(a\) and \(b\) that we found:
\begin{equation}
y=9x-1.
\end{equation}

# data[].attributes.body
Given \(f\) and \(c\in \mathbb{R}\), find the slope of the tangent to the graph of \(f\) at the point \((c,f(c))\) and then find the equation of the mentioned tangent.

# data[].attributes.body
I will solve it with the modified values: \(f(x)=6x^6-10x^4+3\).
The derivative is \(f'(x)=36x^5-40x^3\).
We factor it: \(f'(x)=4x^3(9x^2-10)\). Now, as always, in order to study the sign of a product, study first the sign of its factors.
Here, \(4x^3\) is negative if and only if \(x<0\) and positive if and only if \(x>0\).
What about \(9x^2-10\)? This a quadratic; in this present example, it is parabola with a 'cup' form and two real roots, so it is negative strictly between its roots and positive strictly outside of them. The roots are \(\pm\sqrt{10/9}\), so the quadratic \(9x^2-10\) is negative if and only if \(-\sqrt{10/9}<x<\sqrt{10/9}\) and positive if and only if \(x\in \mathbb{R}\setminus [-\sqrt{10/9},\sqrt{10/9}]\) .
We now combine our information, the product of two factors is negative if and only if the factors are of opposite sign, hence, \(f'(x)<0\) if and only if \begin{equation}
x\in (-\infty,-\sqrt{10/9}) \cup (0,\sqrt{10/9})
\end{equation}

# data[].attributes.body
I will solve the question, with the modified values:\begin{equation}f(x)=\left\{\begin{array}{cr}-3-x^3,& x\leq -1\\
Bx-C,& x>-1\end{array}\right.
\end{equation}
The first thing to notice is that \(f\) is piecewise defined as polynomials on \((-\infty,-1]\) and \((-1,\infty)\) respectively. It implies that \(f\) is, with whatever fixed values of \(B\) and \(C\), continuous and differentiable on \((-\infty,-1)\) and \((-1,\infty)\). We therefore only have to ensure continuity and differentiability at \(-1\), and this will be by finding the right values of \(B\) and \(C\). To find these values, we need to derive equations involving \(B\) and \(C\) and then solve these equations for \(B\) and \(C\). I derive the equations below. One comes from the continuity condition, the other one from the differentiability condition.

Continuity
We first need to ensure continuity of \(f\) at \(x=-1\), that is to say, the following equations must be true:
\begin{equation}
\lim_{x\to -1^-}f(x)=f(-1)=\lim_{x\to  -1^+}f(x).
\end{equation}
Let's compute the three terms above:
\begin{align}
\lim_{x\to -1^-}f(x)&=-3-(-1)^3=-2 \qquad \text{(make sure you can do this)}\\
\lim_{x\to  -1^+}f(x)&=B(-1)-C=-B-C\\
f(-1)&=-3-(-1)^3=-2.
\end{align}
We can now rewrite the continuity equation with specific values:
\begin{equation}
-B-C=-2. \qquad \text{(continuity equation)}
\end{equation}
Differentiability
To ensure differentiability of \(f\) at \(x=-1\), the following equations must hold:
\begin{equation}
\lim_{x\to -1^-}f'(x)=\lim_{x\to  -1^+}f'(x).
\end{equation}
On \((-\infty,-1)\), \(f'(x)=-3x^2\) and on \((-1,\infty)\), \(f'(x)=B\) (again, make sure you know how to find this). We may hence compute the one-sided limits for the derivative of \(f\):
\begin{align}
\lim_{x\to -1^-}f'(x)&=-3(-1)^2=-3 \\
\lim_{x\to  -1^+}f(x)&=B.
\end{align}
We can now rewrite the differentiability equation with specific values:
\begin{equation}
B=-3. \qquad \text{(differentiability equation)}
\end{equation}
Concluding
The differentiability equation gave us a value for \(B\). With the continuity equation we may hence find the value for \(C\):
\begin{equation}
-(-3)-C=-2 \Rightarrow C=5.
\end{equation}
The required values for \(B\) and \(C\) are \(B=-3, C=5\).

Ask about anything that is unclear.