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AP Calculus AB : Derivatives

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Example Questions

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

$\begin{align*}&x&f(x)\\&2&-7\\&3&39\\&4&-11\\&5&27\\&\text{Using the above table, make a forward derivative approximation at }x=3\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(4)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&f'(3)\approx\frac{-11-(-11)}{1}\\&f'(3)\approx-50\end{align*}$

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Example Question #482 : Derivatives

$\begin{align*}&x&f(x)\\&2&45\\&4&-17\\&6&17\\&8&-6\\&10&34\\&\text{Using the above table, make a forward derivative approximation at }x=8\end{align*}$

Possible Answers:

$-\frac{23}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(8)\text{ and }f(10)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(8)\approx\frac{34-(34)}{2}\\&f'(8)\approx20\end{align*}$

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Example Question #77 : Derivative At A Point

$\begin{align*}&x&f(x)\\&3&9\\&6&-35\\&9&-30\\&12&-9\\&15&25\\&18&33\\&21&29\\&24&-18\\&\text{Approximate a forward derivative at }x=12\end{align*}$

Possible Answers:

$\frac{17}{3}$

$\frac{34}{3}$

Correct answer:

$\frac{34}{3}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(12)\text{ and }f(15)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&f'(12)\approx\frac{25-(25)}{3}\\&f'(12)\approx\frac{34}{3}\end{align*}$

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Example Question #481 : Derivatives

$\begin{align*}&\text{Approximate a forward derivative at }x=0\\&\text{Using the following table:}\\&x&f(x)\\&0&-31\\&4&-1\\&8&-36\end{align*}$

Possible Answers:

$\frac{15}{4}$

$-\frac{35}{4}$

$-\frac{31}{4}$

$\frac{15}{2}$

Correct answer:

$\frac{15}{2}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(4)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(0)\approx\frac{-1-(-1)}{4}\\&f'(0)\approx\frac{15}{2}\end{align*}$

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Example Question #81 : Derivative At A Point

$\begin{align*}&x&f(x)\\&1&32\\&5&38\\&9&49\\&13&-50\\&17&37\\&21&11\\&25&49\\&29&3\\&\text{Approximate a forward derivative at }x=13\end{align*}$

Possible Answers:

$\frac{87}{4}$

$-\frac{25}{2}$

$-\frac{99}{4}$

$\frac{87}{8}$

Correct answer:

$\frac{87}{4}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(13)\text{ and }f(17)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(13)\approx\frac{37-(37)}{4}\\&f'(13)\approx\frac{87}{4}\end{align*}$

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Example Question #482 : Derivatives

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

Possible Answers:

$-\frac{37}{5}$

$\frac{37}{5}$

Correct answer:

Explanation:

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

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Example Question #483 : Derivatives

$\begin{align*}&x&f(x)\\&-3&12\\&1&-14\\&5&-45\\&9&-1\\&13&-31\\&17&-38\\&\text{Approximate a backward derivative at }x=1\end{align*}$

Possible Answers:

$\frac{13}{2}$

$-\frac{1}{2}$

$\frac{1}{2}$

$-\frac{13}{2}$

Correct answer:

$-\frac{13}{2}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(1)\text{ and }f(-3)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(1)\approx\frac{-14-(12)}{4}\\&f'(1)\approx-\frac{13}{2}\end{align*}$

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Example Question #484 : Derivatives

$\begin{align*}&x&f(x)\\&-2&4\\&2&20\\&6&0\\&\text{Approximate a backward derivative at }x=2\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(2)\text{ and }f(-2)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(2)\approx\frac{20-(4)}{4}\\&f'(2)\approx4\end{align*}$

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Example Question #485 : Derivatives

$\begin{align*}&x&f(x)\\&3&-23\\&8&-29\\&13&7\\&18&14\\&23&-8\\&\text{Approximate a backward derivative at }x=23\end{align*}$

Possible Answers:

$-\frac{6}{5}$

$\frac{22}{5}$

$\frac{6}{5}$

$-\frac{22}{5}$

Correct answer:

$-\frac{22}{5}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(23)\text{ and }f(18)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&f'(23)\approx\frac{-8-(14)}{5}\\&f'(23)\approx-\frac{22}{5}\end{align*}$

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Example Question #486 : Derivatives

$\begin{align*}&x&f(x)\\&-2&12\\&-1&7\\&0&-45\\&\text{Using the above table, make a backward derivative approximation at }x=0\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(0)\text{ and }f(-1)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&f'(0)\approx\frac{-45-(7)}{1}\\&f'(0)\approx-52\end{align*}$

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

Example Question #482 : Derivatives

Example Question #77 : Derivative At A Point

Example Question #481 : Derivatives

Example Question #81 : Derivative At A Point

Example Question #482 : Derivatives

Example Question #483 : Derivatives

Example Question #484 : Derivatives

Example Question #485 : Derivatives

Example Question #486 : Derivatives

All AP Calculus AB Resources

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