AP Calculus AB - Derivatives

Find

$f(x)=\tan(e^{x^2})$

$f'(x)=2xe^{x^2}\sec^2(e^{x^2})$

$f'(x)=\sec^2(e^{x^2})$

$f'(x)=\tan(2xe^{x^2})$

$f'(x)=e^{x^2}\sec^2(e^{x^2})$

$f'(x)=x^2e^{x^2-1}\sec(e^{x^2})$

Explanation

We are going to use three rules along with the chain rule:

$\frac{\mathrm{d} }{\mathrm{d} x}\tan(x)=\sec^2(x)$

$\frac{\mathrm{d} }{\mathrm{d} x}e^x=e^x$

$\frac{\mathrm{d} }{\mathrm{d} x}x^a=ax^{a-1}$

So then, using our first rule and the chain rule

$\frac{\mathrm{d} }{\mathrm{d} x}\tan(e^{x^2})=\sec^2(e^{x^2})\cdot \frac{\mathrm{d} }{\mathrm{d} x}e^{x^2}$

then using our second rule and chain rule

$=\sec^2(e^{x^2})e^{x^2}\cdot \frac{\mathrm{d} }{\mathrm{d} x}x^2$

then using our third rule (no chain rule this time)

$=\sec^2(e^{x^2})e^{x^2}2x$

Then we rearrange the equation for simplification,

$=2xe^{x^2}\sec^2(e^{x^2})$

$g(x)=x^{3}$

and

$h(x)=7x^{2}+3x$

Find the derivative of

None of these answers

Explanation

The chain rule of the derivative always deals with the composition of two or more functions.

In this case we can identify two,

and

So is the composition of these two such that:

With chain rule, you always start on the outermost function and work your way inward, which in this case is:

Always the derivative of the outermost evaluated at the inner, multiplied by the derivative of the inner.

$f(x) = \cos (\sqrt{x^{2}- 2x} )$

Find .

$f'(x) =- \frac{ ( x-1) \sin \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) =- \frac{ ( x-1) \cos \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) = \frac{ ( x-1) \sin \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) = \frac{ ( x-1) \cos \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) = \frac{ ( x-1) \csc \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

Explanation

Let $u(x) = x^{2}- 2x$

Then

$f(x) = \cos (\sqrt{x^{2}- 2x} )$

can be rewritten as

$f(x) = \cos (\sqrt{u } )$

Let $v (u) = \sqrt{u} =u^{\frac{1}{2}}$

The function can now be rewritten as

$f(x) = \cos v$

Applying the chain rule twice:

$\frac{\mathrm{d} f}{\mathrm{d} x} = \frac{\mathrm{d} f}{\mathrm{d} v} \cdot \frac{\mathrm{d} v}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin v \cdot \frac{1}{2} u ^{-\frac{1}{2}} \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin u^{\frac{1}{2}} \cdot \frac{1}{2} u ^{-\frac{1}{2}} \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin \sqrt{u} \cdot \frac{1}{2\sqrt{u} } \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin \sqrt{x^{2}- 2x} \cdot \frac{1}{2\sqrt{x^{2}- 2x} } \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} = \frac{-\sin \sqrt{x^{2}- 2x} \cdot (2x-2)}{2\sqrt{x^{2}- 2x} }$

$\frac{\mathrm{d} f}{\mathrm{d} x} =- \frac{ ( x-1) \sin \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$\begin{align*}&t&p(t)\&2&24\&3&6\&4&-32\&5&10\&6&-20\&7&-37\&8&-29\&9&40\&\text{The above table lists a particle's postion at different times.}\&\text{Approximate its velocity with a forward difference at }t=2\end{align*}$

Explanation

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\&\text{we do not have a position function, we cannot take the derivative;}\&\text{we must use an approximation. To find a forward derivative approximation,}\&\text{we consider function two values: one at the point of interest}\&\text{and another succeeding it, and finally that distance between}\&\text{the two. Mathematically, this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(2)\text{ and }f(3)\&\text{Applying the formula above, with }h=1\text{, we find:}\&v(2)\approx\frac{6-(24)}{1}\&v(2)\approx-18\end{align*}$

$\begin{align*}&\text{Approximate the acceleration of a particle using}\&\text{a central difference approach at }t=6\&\text{Using the following table of positions:}\&t&p(t)\&2&26\&6&-23\&10&18\&14&16\&18&-34\end{align*}$

$\frac{45}{8}$

$-\frac{49}{4}$

$-\frac{1}{2}$

Explanation

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\&\text{words, you take the derivative of a position function twice}\&\text{to define it. In this problem, we’re not given a function, but}\&\text{ a table of times and positions, requiring an approximation,}\&\text{a 2nd order central difference. This has the form:}\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{This approximation is actually born from two types of first}\&\text{order approximations: the forward and backward differences,}\&\text{which have the forms:}\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\&\text{The difference of these two, divided by our increment, gives}\&\text{a derivative of a derivative.}\&\text{Applying the formula above, with }h=4\text{, we find:}\&a(6)\approx\frac{18+(26)-2(-23)}{4^2}\&a(6)\approx\frac{45}{8}\end{align*}$

$\begin{align*}&x&f(x)\&3&27\&8&8\&13&43\&18&8\&23&-49\&28&-38\&33&37\&\text{Approximate a backward derivative at }x=33\end{align*}$

$-\frac{37}{5}$

$\frac{37}{5}$

Explanation

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another preceding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x)-f(x-h)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(33)\text{ and }f(28)\&\text{Applying the formula above, with }h=5\text{, we find:}\&f'(33)\approx\frac{37-(-38)}{5}\&f'(33)\approx15\end{align*}$

$\begin{align*}&\text{Approximate a forward derivative at }x=4\&\text{Using the following table:}\&x&&f(x)\&1&&5\&4&&-8\&7&&15\end{align*}$

$\frac{23}{3}$

$-\frac{13}{3}$

$\frac{23}{6}$

$-\frac{8}{3}$

Explanation

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\&\text{two values: one at the point of interest and another succeeding}\&\text{it, and finally that distance between the two. Mathematically,}\&\text{this follows the form:}\&f'(x)=\frac{f(x+h)-f(x)}{h}\&\text{Where h is the size of the increment between points on the x-axis.}\&\text{In approximating a derivative, it is generally prudent to minimize}\&\text{the distance between function values. In other words, it’s a}\&\text{good idea to look at two adjacent points. For our problem, that’d}\&\text{be:}\&f(4)\text{ and }f(7)\&\text{Applying the formula above, with }h=3\text{, we find:}\&f'(4)\approx\frac{15-(15)}{3}\&f'(4)\approx\frac{23}{3}\end{align*}$

Determine the rate of change of the angle opposite the base of a right triangle -whose length is increasing at a rate of 1 inch per minute, and whose height is a constant 2 inches - when the area of the triangle is 2 square inches.

$\frac{1}{\sqrt{2}}$ radians per minute

$\frac{1}{2\sqrt{2}}$ radians per minute

$\frac{1}{4\sqrt{2}}$ radians per minute

$\frac{1}{4}$ radians per minute

Explanation

To determine the rate of the change of the angle opposite to the base of the given right triangle, we must relate it to the rate of change of the base of the triangle when the triangle is a certain area.

First we must determine the length of the base of the right triangle at the given area:

$A=\frac{bh}{2}$

Now, we must find something that relates the angle opposite of the base to the length of the base and height - the tangent of the angle:

$\tan \theta=\frac{b}{h}$

To find the rate of change of the angle, we take the derivative of both sides with respect to time, keeping in mind that the base of the triangle is dependent on time, while the height is constant:

$\sec^2(\theta)\frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{2}{2}\frac{\mathrm{d} b}{\mathrm{d} t}$

We know the rate of change of the base, and we can find the angle from the sides of the triangle:

$\tan(\theta)=1$

$\theta=\frac{\pi}{4}$

Plugging this and the other known information in and solving for the rate of change of the angle adjacent to the base, we get

$\sec^2(\frac{\pi}{4})\frac{\mathrm{d} \theta}{\mathrm{d} t}=1$

$\frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{1}{\sqrt{2}}$ radians per minute

Find the derivative of the following function at the point .

Explanation

Use the power rule on each term of the polynomial to get the derivative,

Now we plug in

Given that , compute the derivative of the following function

$y=6\sin(xy)+e^y$

$y'=\frac{6y\cos(xy)}{1-6x\cos(xy)-e^y}$

$y'=\frac{6y\sin(xy)}{1-6\cos(xy)-e^y}$

$y'=\frac{6y\cos(xy)}{1+6\cos(xy)-e^y}$

$y'=\frac{6y\cos(xy)}{1-6\cos(xy)-ye^y}$

Explanation

To find the derivative of the function, we use implicit differentiation, which is an application of the chain rule. We use this because , and any derivative with respect to is $\frac{\mathrm{d} y}{\mathrm{d} x}$ (or ).