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AP Calculus AB : Derivatives

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Example Questions

Example Question #471 : Derivatives

Use the chain rule to differentiate the following function: $h(x)=sin(x^{2}e^{x})$

Possible Answers:

$h^{'}(x)=sin(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=-cos(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=-sin(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

Correct answer:

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

Explanation:

$h(x)=sin(x^{2}e^{x})$

By the chain rule: $h^{'}(x)=cos(x^{2}e^{x})\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$

Differentiate $(x^{2}e^{x})$ using the product rule: $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})e^{x}+x^{2}\frac{\mathrm{d} }{\mathrm{d} x}e^{x}=2xe^{x}+x^{2}e^{x}$

Substitute this derivative for $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$ in the first equation: $h^{'}(x)=cos(x^{2}e^{x})(2xe^{x}+x^{2}e^{x})$

Factor the equation: $h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

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Example Question #771 : Ap Calculus Ab

f(p)=cos(2p-1)-(2p^3+4p)

Find f'(p) .

Possible Answers:

f'(p)=cos(2p-1)+sin(2p-1)-6p^2+4

f'(p)=-2cos(2p-1)(p^3+3p^2+2p+2)

f'(p)=sin(2p-1)-sin(2p^3+4p)

f'(p) is undefined.

f'(p)=-2(sin(2p-1)+3p^2+2)

Correct answer:

f'(p)=-2(sin(2p-1)+3p^2+2)

Explanation:

$\frac{d}{dx}[cos(2p-1)]=2 \cdot -sin(2p-1)=-2sin(2p-1)$

$\frac{d}{dx}[2p^3+4p]=6p^2+4$

Therefore:

f'(p)=-2sin(2p-1)-(6p^2+4)

=-2sin(2p-1)-6p^2-4

=-2(sin(2p-1)+3p^2+2)

f'(p)=-2(3p^2+sin(2p-1)+2)

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Example Question #821 : Ap Calculus Ab

Which of the following functions contains a removeable discontinuity?

Possible Answers:

$f(x)=\frac{x+1}{1+x^{2}}$

$f(x)=1-(\ln x)^{2}$

$f(x)=x^{2}\sin x^{2}$

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

$f(x)=\frac{1+x^{3}}{1+x}$

Correct answer:

$f(x)=\frac{1+x^{3}}{1+x}$

Explanation:

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at x=a , then the limit as approaches exists, but the value of f(a) does not.

For example, the function $f(x)=\frac{1+x^3}{1+x}$ contains a removeable discontinuity at x=-1 . Notice that we could simplify f(x) as follows:

$f(x)=\frac{1+x^3}{1+x}=\frac{(1+x)(x^2-x+1)}{1+x}=x^{2}-x+1$ , where $x\neq -1$ .

Thus, we could say that $\lim_{x\rightarrow -1}\frac{1+x^3}{1+x}=\lim_{x\rightarrow -1}x^2-x+1=(-1)^2-(-1)+1=3$ .

As we can see, the limit of f(x) exists at x=-1 , even though f(-1) is undefined.

What this means is that f(x) will look just like the parabola with the equation $x^{2}-x+1$ EXCEPT when x=-1 , where there will be a hole in the graph. However, if we were to just define f(-1)=3 , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at x=-1 .

The functions

$f(x)=\frac{x}{1-x^{2}}$ , and

$f(x)=1-(\ln x)^{2}$

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

$f(x)=x^{2}\sin x^{2}$ and $f(x)=\frac{x+1}{1+x^{2}}$ are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

$f(x)=\frac{1+x^{3}}{1+x}$ .

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Example Question #1 : Approximate Rate Of Change From Graphs And Tables Of Values

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

Possible Answers:

$\frac{23}{3}$

$\frac{23}{6}$

$-\frac{8}{3}$

$-\frac{13}{3}$

Correct answer:

$\frac{23}{3}$

Explanation:

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

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Example Question #71 : Derivative At A Point

$\begin{align*}&x&f(x)\\&1&-27\\&3&-38\\&5&11\\&7&-5\\&9&-4\\&11&16\\&13&27\\&15&-15\\&\text{Using the above table, make a forward derivative approximation at }x=9\end{align*}$

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(9)\text{ and }f(11)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(9)\approx\frac{16-(16)}{2}\\&f'(9)\approx10\end{align*}$

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Example Question #72 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=7\\&\text{Using the following table:}\\&x&f(x)\\&1&8\\&3&4\\&5&37\\&7&-24\\&9&-18\\&11&-38\\&13&44\end{align*}$

Possible Answers:

$-\frac{61}{2}$

$\frac{3}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(7)\text{ and }f(9)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(7)\approx\frac{-18-(-18)}{2}\\&f'(7)\approx3\end{align*}$

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Example Question #73 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=0\\&\text{Using the following table:}\\&x&f(x)\\&0&22\\&4&2\\&8&50\\&12&-28\\&16&-40\\&20&-39\end{align*}$

Possible Answers:

$-\frac{5}{2}$

$\frac{11}{2}$

$-\frac{39}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(4)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(0)\approx\frac{2-(2)}{4}\\&f'(0)\approx-5\end{align*}$

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Example Question #74 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=9\\&\text{Using the following table:}\\&x&f(x)\\&1&27\\&5&44\\&9&48\\&13&-31\\&17&-36\end{align*}$

Possible Answers:

$-\frac{79}{4}$

$-\frac{79}{8}$

Correct answer:

$-\frac{79}{4}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(9)\text{ and }f(13)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(9)\approx\frac{-31-(-31)}{4}\\&f'(9)\approx-\frac{79}{4}\end{align*}$

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Example Question #75 : Derivative At A Point

$\begin{align*}&x&f(x)\\&0&-11\\&5&17\\&10&24\\&15&2\\&20&-15\\&25&-35\\&\text{Approximate a forward derivative at }x=10\end{align*}$

Possible Answers:

$-\frac{11}{5}$

$-\frac{22}{5}$

$\frac{24}{5}$

$\frac{7}{5}$

Correct answer:

$-\frac{22}{5}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(10)\text{ and }f(15)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&f'(10)\approx\frac{2-(2)}{5}\\&f'(10)\approx-\frac{22}{5}\end{align*}$

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Example Question #76 : Derivative At A Point

$\begin{align*}&\text{Approximate a forward derivative at }x=0\\&\text{Using the following table:}\\&x&f(x)\\&0&19\\&2&-14\\&4&24\\&6&-11\\&8&19\\&10&21\\&12&-6\end{align*}$

Possible Answers:

$\frac{19}{2}$

$-\frac{33}{2}$

$-\frac{33}{4}$

Correct answer:

$-\frac{33}{2}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(2)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(0)\approx\frac{-14-(-14)}{2}\\&f'(0)\approx-\frac{33}{2}\end{align*}$

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #471 : Derivatives

Example Question #771 : Ap Calculus Ab

Example Question #821 : Ap Calculus Ab

Example Question #1 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #71 : Derivative At A Point

Example Question #72 : Derivative At A Point

Example Question #73 : Derivative At A Point

Example Question #74 : Derivative At A Point

Example Question #75 : Derivative At A Point

Example Question #76 : Derivative At A Point

All AP Calculus AB Resources

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