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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #214 : Computation Of The Derivative

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-119tan(6x)e^{(17x)}\end{align*}$

Possible Answers:

$12138tan(6x)e^{(17x)} - 2023e^{(17x)}\cdot (6tan(6x)^{2} + 6)$

$289e^{(17x)} - 42tan(6x)^{2} - 42$

$-2023e^{(17x)}\cdot (6tan(6x)^{2} + 6)$

$- 119e^{(17x)}\cdot (6tan(6x)^{2} + 6) - 2023tan(6x)e^{(17x)}$

Correct answer:

$- 119e^{(17x)}\cdot (6tan(6x)^{2} + 6) - 2023tan(6x)e^{(17x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }e^{(17x)}\\&\text{and the second being: }-tan(6x)\\&\text{With the product scaled by }119\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[e^{ax}]=ae^{ax}dx\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&\text{From the above we can continue:}\\&u=e^{(17x)};du=17e^{(17x)}\\&v=-tan(6x);dv=- 6tan(6x)^{2} - 6\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 119e^{(17x)}\cdot (6tan(6x)^{2} + 6) - 2023tan(6x)e^{(17x)}\end{align*}$

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Example Question #215 : Computation Of The Derivative

$\begin{align*}&\text{Calculate the derivative of the function}\\&-\frac{(272cos(12x))}{3^{(12x)}}\end{align*}$

Possible Answers:

$\frac{(3264sin(12x))}{3^{(12x)}}+\frac{ (3264cos(12x)ln(3))}{3^{(12x)}}$

$\frac{(204ln(3))}{3^{(12x)}}- 192sin(12x)$

$\frac{(39168cos(12x)ln(3))}{3^{(12x)}}-\frac{ (39168sin(12x))}{3^{(12x)}}$

$-\frac{(39168sin(12x)ln(3))}{3^{(12x)}}$

Correct answer:

$\frac{(3264sin(12x))}{3^{(12x)}}+\frac{ (3264cos(12x)ln(3))}{3^{(12x)}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }\frac{1}{3^{(12x)}}\\&\text{and the second being: }cos(12x)\\&\text{With the product scaled by }-272\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&d[cos(ax)]=-asin(ax)dx\\&\text{From the above we can continue:}\\&u=\frac{1}{3^{(12x)}};du=-\frac{(12ln(3))}{3^{(12x)}}\\&v=cos(12x);dv=-12sin(12x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(3264sin(12x))}{3^{(12x)}}+\frac{ (3264cos(12x)ln(3))}{3^{(12x)}}\end{align*}$

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Example Question #216 : Computation Of The Derivative

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-112x^{17}ln(-4x)\end{align*}$

Possible Answers:

$119x^{16} -\frac{ 16}{x}$

$-1904x^{15}$

$- 1904x^{16}ln(-4x) - 112x^{16}$

$7616x^{16}ln(-4x) - 1904x^{16}$

Correct answer:

$- 1904x^{16}ln(-4x) - 112x^{16}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(-4x)\\&\text{and the second being: }x^{17}\\&\text{With the product scaled by }-112\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=ln(-4x);du=\frac{1}{x}\\&v=x^{17};dv=17x^{16}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 1904x^{16}ln(-4x) - 112x^{16}\end{align*}$

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Example Question #221 : Computation Of The Derivative

$\begin{align*}&\text{Calculate the derivative of the function}\\&130cos(6x)e^{(-14x)}\end{align*}$

Possible Answers:

$- 1820cos(6x)e^{(-14x)} - 780sin(6x)e^{(-14x)}$

$10920cos(6x)e^{(-14x)} + 10920sin(6x)e^{(-14x)}$

$10920sin(6x)e^{(-14x)}$

$- 182e^{(-14x)} - 60sin(6x)$

Correct answer:

$- 1820cos(6x)e^{(-14x)} - 780sin(6x)e^{(-14x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }e^{(-14x)}\\&\text{and the second being: }cos(6x)\\&\text{With the product scaled by }130\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[e^{ax}]=ae^{ax}dx\\&d[cos(ax)]=-asin(ax)dx\\&\text{From the above we can continue:}\\&u=e^{(-14x)};du=-14e^{(-14x)}\\&v=cos(6x);dv=-6sin(6x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 1820cos(6x)e^{(-14x)} - 780sin(6x)e^{(-14x)}\end{align*}$

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Example Question #11 : Derivative Rules For Sums, Products, And Quotients Of Functions

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-\frac{(220cos(3x))}{x^{16}}\end{align*}$

Possible Answers:

$33sin(3x) -\frac{ 320}{x^{17}}$

$\frac{(3520cos(3x))}{x^{17}}+\frac{ (660sin(3x))}{x^{16}}$

$-\frac{(10560sin(3x))}{x^{17}}$

$\frac{(10560cos(3x))}{x^{17}}-\frac{ (10560sin(3x))}{x^{16}}$

Correct answer:

$\frac{(3520cos(3x))}{x^{17}}+\frac{ (660sin(3x))}{x^{16}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }cos(3x)\\&\text{and the second being: }\frac{1}{x^{16}}\\&\text{With the product scaled by }-220\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[cos(ax)]=-asin(ax)dx\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=cos(3x);du=-3sin(3x)\\&v=\frac{1}{x^{16}};dv=-\frac{16}{x^{17}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(3520cos(3x))}{x^{17}}+\frac{ (660sin(3x))}{x^{16}}\end{align*}$

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Example Question #223 : Computation Of The Derivative

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=32x^{17}tan(4x)\end{align*}$

Possible Answers:

$16tan(4x)^{2} + 136x^{16} + 16$

$544x^{16}\cdot (4tan(4x)^{2} + 4)$

$544x^{16}tan(4x) + 32x^{17}\cdot (4tan(4x)^{2} + 4)$

$2176x^{16}tan(4x) + 544x^{17}\cdot (4tan(4x)^{2} + 4)$

Correct answer:

$544x^{16}tan(4x) + 32x^{17}\cdot (4tan(4x)^{2} + 4)$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }x^{17}\\&\text{and the second being: }tan(4x)\\&\text{With the product scaled by }32\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[x^a]=ax^{a-1}dx\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&\text{From the above we can continue:}\\&u=x^{17};du=17x^{16}\\&v=tan(4x);dv=4tan(4x)^{2} + 4\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&544x^{16}tan(4x) + 32x^{17}\cdot (4tan(4x)^{2} + 4)\end{align*}$

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Example Question #224 : Computation Of The Derivative

Find f'(x) for the following f(x).

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

Possible Answers:

$f'(x)=2\cdot (cos(x)+sin(x))$

$f'(x)=4\cdot e^{x}\cdot sin(x)$

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

$f'(x)= e^{x}(cos(x)+sin(x))^{2}$

$f'(x)=2\cdot e^{x}(cos(x)+sin(x))$

Correct answer:

$f'(x)=2\cdot e^{x}(cos(x)+sin(x))$

Explanation:

Since the function f(x) is a product of two functions, 2sin(x) and e^x, use the product rule to take this derivative.

Recall the product rule where u and v are two separate functions:

(uv)'=u'v+uv'

Let's define u and v as follows:

u=2sin(x)

$v=e^{x}$

Take the derivative of u and v:

u'=2cos(x)

$v'=e^{x}$

Now that we've found values of u, u', v, and v' we can substitute them into the product formula for the final answer.

$f'(x)=(uv)'=u'v+uv'=2cos(x)\cdot e^{x}+2sin(x)\cdot e^{x}$

For the answer in final form, factor out 2*e^x.

$f'(x)=2\cdot e^{x}(cos(x)+sin(x))$

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Example Question #225 : Computation Of The Derivative

Find f'(x) for the following f(x).

$f(x)=tan(x)\cdot ln(x)$

Possible Answers:

$f'(x)=\frac{tan(x)}{x}+ln(x)\cdot cot^{2}x$

$f'(x)=tan(x)\cdot x+ln(x)\cdot sec^{2}x$

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

$f'(x)=\frac{tan(x)}{x}+ln(x)\cdot sec^{2}x$

$f'(x)=\frac{tan(x)}{x}+sec^{2}x$

Correct answer:

$f'(x)=\frac{tan(x)}{x}+ln(x)\cdot sec^{2}x$

Explanation:

Since the function f(x) is a product of two functions, tan(x) and ln(x), use the product rule to take this derivative.

Recall the product rule where u and v are two separate functions:

(uv)'=u'v+uv'

Let's define u and v as follows:

u=tan(x)

v=ln(x)

Take the derivative of u and v:

$u'=sec^{2}x$

$v'=\frac{1}{x}$

Now that we've found values of u, u', v, and v' we can substitute them into the product formula for the final answer.

$f'(x)=(uv)'=u'v+uv'=\frac{tan(x)}{x}+ln(x)\cdot sec^{2}x$

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Example Question #226 : Computation Of The Derivative

Find f'(x) for the following f(x).

$f(x)=\pi e^{x}\cdot ln(x)$

Possible Answers:

$f'(x)=\pi \cdot e^{x}\left [ \frac{1}{x}+e^{x}ln(x)\right ]$

$f'(x)=\pi \cdot e^{x}\left [ \frac{1}{x}+ln(x)\right ]$

$f'(x)=e^{x}\left [ \frac{1}{x}+ln(x)\right ]$

$f'(x)=\pi \cdot e^{x}\left [ \frac{1}{x}-ln(x)\right ]$

$f'(x)=\pi \cdot e^{x}\left [ \frac{1}{x}+x\cdot ln(x)\right ]$

Correct answer:

$f'(x)=\pi \cdot e^{x}\left [ \frac{1}{x}+ln(x)\right ]$

Explanation:

Since the function f(x) is a product of two functions, pi*e^x and ln(x), use the product rule to take this derivative.

Recall the product rule where u and v are two separate functions:

(uv)'=u'v+uv'

Let's define u and v as follows:

$u=\pi e^{x}$

v=ln(x)

Take the derivative of u and v:

$u'=\pi e^{x}$

$v'=\frac{1}{x}$

Now that we've found values of u, u', v, and v' we can substitute them into the product formula for the final answer.

$f'(x)=(uv)'=u'v+uv'=\pi e^{x}\cdot ln(x)+\frac{\pi e^{x}}{x}$

Factor out the pi*e^x term to get the answer into final form.

$f'(x)=\pi \cdot e^{x}\left [ \frac{1}{x}+ln(x)\right ]$

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Example Question #221 : Computation Of The Derivative

Find f'(x) for the following f(x).

$f(x)=-cos(x)+ x^{6}sin(x)$

Possible Answers:

$f'(x)=sin(x)(1+x^6)+x^{6}cos(x)$

$f'(x)=sin(x)(1+6x^5)+x^{6}cos(x)$

$f'(x)=cos(x)(1+6x^5)+x^{6}sin(x)$

$f'(x)=sin(x)(1+6x^5)+6x^{5}cos(x)$

$f'(x)=sin(x)(1+6x^5)-x^{6}cos(x)$

Correct answer:

$f'(x)=sin(x)(1+6x^5)+x^{6}cos(x)$

Explanation:

The function f(x) is a sum of two functions, -cos(x) and x^6*sin(x). When functions are added together, treat each one separately to take it's derivative. First take the derivative of -cos(x):

$\frac{d}{dx}-cos(x)=sin(x)$

The second part of the function, x^6*sin(x) is a product of two functions so use the product rule to take the derivative. Recall the product rule where u and v are two separate functions:

(uv)'=u'v+uv'

Let's define u and v as follows:

$u=x^{6}$

v=sin(x)

Take the derivative of u and v:

$u'=6x^{5}$

v'=cos(x)

Now that we've found values of u, u', v, and v' substitute them into the product rule formula.

$(uv)'=u'v+uv'=6x^{5}sin(x)+x^{6}cos(x)$

For the final answer add (uv)' to the derivative of -cos(x).

$f'(x)=sin(x)+6x^{5}sin(x)+x^{6}cos(x)$

Simplify into final form by factoring out sin(x):

$f'(x)=sin(x)(1+6x^5)+x^{6}cos(x)$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #214 : Computation Of The Derivative

Example Question #215 : Computation Of The Derivative

Example Question #216 : Computation Of The Derivative

Example Question #221 : Computation Of The Derivative

Example Question #11 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #223 : Computation Of The Derivative

Example Question #224 : Computation Of The Derivative

Example Question #225 : Computation Of The Derivative

Example Question #226 : Computation Of The Derivative

Example Question #221 : Computation Of The Derivative

All AP Calculus AB Resources

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