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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for 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 #228 : Computation Of The Derivative

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

$f(x)=4^{x}cot(x)$

Possible Answers:

$f'(x)=ln(4)\cdot cot(x)-4^{x}csc^{2}x$

$f'(x)=ln(4)\cdot 4^{x}\cdot cot(x)-4^{x}sec^{2}x$

$f'(x)=4^{x}ln(4)cot(x)-4^{x}csc^{2}x$

$f'(x)= 4^{x}\cdot cot(x)-4^{x}csc^{2}x$

$f'(x)=ln(4)\cdot 4^{x}\cdot cot(x)+4^{x}csc^{2}x$

Correct answer:

$f'(x)=4^{x}ln(4)cot(x)-4^{x}csc^{2}x$

Explanation:

Since the function f(x) is a product of two functions, 4^x and cot(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:

$u=4^{x}$

v=cot(x)

Take the derivative of u and v:

$u'=4^{x}ln(4)$

$v'=-csc^{2}x$

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

$f'(x)=(uv)'=u'v+uv'=4^{x}ln(4) cot(x)-4^{x}csc^{2}x$

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Example Question #229 : Computation Of The Derivative

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

$f(x)=\frac{tanx}{e^{x}}$

Possible Answers:

$f'(x)=\frac{sec^{2}x-2\cdot tanx}{e^{x}}$

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

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

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

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

Correct answer:

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

Explanation:

Since the function f(x) is a quotient of two separate functions, tan(x) and e^x, use the quotient rule to take the derivative.

Recall the quotient rule for two functions u and v:

$(\frac{u}{v})^{'}=\frac{u'v-uv'}{v^{2}}$

Let's define u and v for this function as follows:

u=tan(x)

$v=e^{x}$

Take the derivative of u and v:

$u'=sec^{2}x$

$v'=e^{x}$

Now that we've defined u, u', v, and v' we can substitute them into the quotient rule formula.

$(\frac{u}{v})^{'}=\frac{u'v-uv'}{v^{2}}=\frac{sec^{2}(x) e^{x}-tan(x)e^{x}}{e^{2x}}$

Divide the numerator and denominator by e^x for the answer in its final form.

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

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Example Question #230 : Computation Of The Derivative

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

$f(x)=\frac{x^{4}}{x^{3}+1}$

Possible Answers:

$f'(x)=\frac{7x^{6}+4x^{3}}{(x^{3}+1)^{3}}$

$f'(x)=\frac{x^{6}-4x^{3}}{(x^{3}+1)^{2}}$

$f'(x)=\frac{x^{6}+4x^{3}}{(x^{3}+1)^{2}}$

$f'(x)=\frac{x^{6}+4x^{3}}{(x^{3}-1)^{2}}$

$f'(x)=\frac{x^{6}+4x^{3}+1}{(x^{3}+1)^{2}}$

Correct answer:

$f'(x)=\frac{x^{6}+4x^{3}}{(x^{3}+1)^{2}}$

Explanation:

Since the function f(x) is a quotient of two separate functions, x^4 and x^3+1, use the quotient rule to take the derivative.

Recall the quotient rule for two functions u and v:

$(\frac{u}{v})^{'}=\frac{u'v-uv'}{v^{2}}$

Let's define u and v for this function as follows:

u=x^4

$v=x^{3}+1$

Take the derivative of u and v:

$u'=4x^{3}$

$v'=3x^{2}$

Now that we've defined u, u', v, and v' we can substitute them into the quotient rule formula.

$(\frac{u}{v})^{'}=\frac{4x^{3}(x^{3}+1)-x^{4}3x^{2}}{(x^{3}+1)^{2}}$

Distribute the 4x^3 and simplify for the final answer.

$f'(x)=\frac{x^{6}+4x^{3}}{(x^{3}+1)^{2}}$

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Example Question #21 : Derivative Rules For Sums, Products, And Quotients Of Functions

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

$f(x)=\frac{10^{x}}{cos(x)}$

Possible Answers:

$f'(x)=\frac{10^{x}(ln(10)cos(x)+sin(x))}{cos^{2}x}$

$f'(x)=\frac{10^{x}(ln(10)cos(x)-sin(x))}{cos^{2}x}$

$f'(x)=\frac{10^{x}(10\cdot cos(x)+sin(x))}{cos^{2}x}$

$f'(x)=\frac{10^{x}(ln(10)cos(x)+sin(x))}{sin^{2}x}$

$f'(x)=\frac{10^{x}(ln(x)cos(x)+sin(x))}{cos^{2}x}$

Correct answer:

$f'(x)=\frac{10^{x}(ln(10)cos(x)+sin(x))}{cos^{2}x}$

Explanation:

Since the function f(x) is a quotient of two separate functions, 10^x and cos(x), use the quotient rule to take the derivative.

Recall the quotient rule for two functions u and v:

$(\frac{u}{v})^{'}=\frac{u'v-uv'}{v^{2}}$

Let's define u and v for this function as follows:

$u=10^{x}$

v=cos(x)

Take the derivative of u and v:

$u'=ln(10)10^{x}$

v'=-sin(x)

Now that we've defined u, u', v, and v' we can substitute them into the quotient rule formula for the final answer.

$(\frac{u}{v})^{'}=\frac{u'v-uv'}{v^{2}}=f'(x)=\frac{10^{x}(ln(10)cos(x)+sin(x))}{cos^{2}x}$

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Example Question #22 : Derivative Rules For Sums, Products, And Quotients Of Functions

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

$f(x)=3x^{7}-x^{10}\cdot sin(x)$

Possible Answers:

$f'(x)=21x^{6}-10x^{9}sin(x)-x^{10}sin(x)$

$f'(x)=21x^{6}+10x^{9}cos(x)-x^{10}sin(x)$ e

$f'(x)=21x^{6}-10x^{9}sin(x)+x^{10}cos(x)$

$f'(x)=21x^{6}-10x^{9}sin(x)-x^{10}cos(x)\cdot 10x^{9}$

$f'(x)=21x^{6}-10x^{9}sin(x)-x^{10}cos(x)$

Correct answer:

$f'(x)=21x^{6}-10x^{9}sin(x)-x^{10}cos(x)$

Explanation:

The function f(x) is a sum of two functions, 3x^7 and -x^10*sin(x). When functions are added together, take the derivative of each one separately. First take the derivative of 3x^7:

$\frac{d}{dx}3x^{7}=21x^{6}$

The second part of the function, -x^10*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:

$u=-x^{10}$

v=sin(x)

Take the derivative of u and v:

$u'=-10x^{9}$

v'=cos(x)

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

$(uv)'=u'v+uv'=-10x^{9}sin(x)-x^{10}cos(x)$

For the final answer add (uv)' to the derivative of 3x^7.

$f'(x)=21x^{6}-10x^{9}sin(x)-x^{10}cos(x)$

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Example Question #23 : Derivative Rules For Sums, Products, And Quotients Of Functions

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

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

Possible Answers:

$f'(x)=\frac{-sinx(x)x^{2}-2xcosx}{(x^{2}+1)^{2}}$

$f'(x)=\frac{-cosx(x^{2}+1)-2xsinx}{(x^{2}+1)^{2}}$

$f'(x)=\frac{-sinx(x)(x^{2}+1)-2xcosx}{(x^{2}+1)^{2}}$

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

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

Correct answer:

$f'(x)=\frac{-sinx(x)(x^{2}+1)-2xcosx}{(x^{2}+1)^{2}}$

Explanation:

Since the function f(x) is a quotient of two separate functions, cos(x) and x^2+1, use the quotient rule to take the derivative.

Recall the quotient rule for two functions u and v:

$(\frac{u}{v})^{'}=\frac{u'v-uv'}{v^{2}}$

Let's define u and v for this function as follows:

u=cosx

$v=x^{2}+1$

Take the derivative of u and v:

u'=-sinx

v'=2x

Now that we've defined u, u', v, and v' we can substitute them into the quotient rule formula for the final answer.

$f'(x)=(\frac{u}{v})^{'}=\frac{-sinx(x^{2}+1)-cosx \cdot 2x}{(x^{2}+1)^{2}}$

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Example Question #31 : Derivative Rules For Sums, Products, And Quotients Of Functions

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

$f(x)=\frac{sinx+cosx}{e^{x}}$

Possible Answers:

$f'(x)=\frac{-2cosx}{e^{x}}$

$f'(x)=\frac{-2sinx}{e^{2x}}$

$f'(x)=\frac{-2e^{x}sinx}{e^{x}}$

$f'(x)=\frac{-2sinx}{e^{x}}$

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

Correct answer:

$f'(x)=\frac{-2sinx}{e^{x}}$

Explanation:

Since the function f(x) is a quotient of two separate functions, sin(x) + cos(x) and e^x, use the quotient rule to take the derivative.

Recall the quotient rule for two functions u and v:

$(\frac{u}{v})^{'}=\frac{u'v-uv'}{v^{2}}$

Let's define u and v for this function as follows:

u=sinx+cosx

$v=e^{x}$

Take the derivative of u and v:

u'=cosx-sinx

$v'=e^{x}$

Now that we've defined u, u', v, and v' we can substitute them into the quotient rule formula.

$f'(x)=(\frac{u}{v})^{'}=\frac{(cosx-sinx)e^{x}-(sinx+cosx)e^{x}}{e^{2x}}$

Factor out e^x from to the top and bottom and simplify into the final form:

$f'(x)=\frac{-2sinx}{e^{x}}$

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Example Question #32 : Derivative Rules For Sums, Products, And Quotients Of Functions

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

$f(x)=5x^{4}\cdot sec(x)$

Possible Answers:

$f'(x)=20x^{3}\cdot sec(x)\cdot tan(x)$

$f'(x)=20x^{3}\cdot sec(x)+5x^{4}\cdot sec(x)\cdot tan(x)$

$f'(x)= sec(x)+5x^{4}$

$f'(x)=20x^{3}+ sec(x)\cdot tan(x)$

$f'(x)=20x^{3}-5x^{4}\cdot sec(x)\cdot tan(x)$

Correct answer:

$f'(x)=20x^{3}\cdot sec(x)+5x^{4}\cdot sec(x)\cdot tan(x)$

Explanation:

Since the function f(x) is a product of two functions, 5x^4 and sec(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:

$u=5x^{4}$

v=sec(x)

Take the derivative of u and v:

$u'=20x^{3}$

$v'=sec(x)\cdot tan(x)$

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

$f'(x)=(uv)'=u'v+uv'=20x^{3}\cdot sec(x)+5x^{4}\cdot sec(x)\cdot tan(x)$

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Example Question #33 : Derivative Rules For Sums, Products, And Quotients Of Functions

Differentiate the function:

$y=x^2\sqrt{3x+1}$

Possible Answers:

$2x\sqrt{3x+1}+\frac{1}{2}(3x+1)$

$= \frac{x(13x+2)}{\sqrt{3x+1}}+1$

$\frac{x}{2\sqrt{3x+1}}$

$3\sqrt{3x+1}+\frac{1}{2}(3x+1)^{-1/2}$

$= \frac{x(15x+4)}{2\sqrt{3x+1}}$

Correct answer:

$= \frac{x(15x+4)}{2\sqrt{3x+1}}$

Explanation:

$y = x^2 \sqrt{3x+1}$

First notice that the function is a product of two functions x^2 and $\sqrt{3x+1}$ . Apply the product rule:

$\frac{dy}{dx}=\sqrt{3x+1}\left(\frac{d}{dx}x^2 \right )+x^2\left(\frac{d}{dx}\sqrt{3x+1} \right )$

_______________________________________________________________

The second term will require the chain rule. Recall that the derivative for a radical function-of-a-function $y=\sqrt{u(x)}$ is given by:

$\underset{Chain-Rule}{ \underbrace{\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}}} =\frac{1}{2\sqrt{u(x)}}u'(x)$

_______________________________________________________________

$\frac{dy}{dx}=\sqrt{3x+1}(2x )+x^2\left(\frac{3}{2\sqrt{3x+1}}\right)$

$=2x\sqrt{3x+1}+\frac{3x^2}{2\sqrt{3x+1}}$

$= \frac{2x\sqrt{3x+1}(2\sqrt{3x+1} )+3x^2}{2\sqrt{3x+1}}$

$= \frac{4x(3x+1)+3x^2}{2\sqrt{3x+1}}$

$= \frac{12x^2+4x+3x^2}{2\sqrt{3x+1}}$

$= \frac{15x^2+4x}{2\sqrt{3x+1}}$

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Example Question #34 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative $\frac{dy}{dx}$

$y= \frac{2x-1}{\sqrt{x^2+4}}$

Possible Answers:

$=\frac{x-1}{x^2+4 }$

$=\frac{x+8}{(x^2+4)^{3/2}}$

$=\frac{x-4}{(x^2+4)^{3/2}}$

$=\frac{2(x-4)}{\sqrt{x^2 +4}}+1$

$=\frac{x+2}{(x^2+4)^{2}}$

Correct answer:

$=\frac{x+8}{(x^2+4)^{3/2}}$

Explanation:

$y= \frac{2x-1}{\sqrt{x^2+4}}$

The derivative can be computed using the quotient rule:

$y = \frac{f(x)}{g(x)}\: \: \: \: \: \:\ \Rightarrow \: \: \: \: \: \:\frac{dy}{dx}=\frac{g(x)f'(x)-g'(x)f(x)}{[g(x)]^2 }$

$\frac{dy}{dx}=\frac{\sqrt{x^2+4}\left[\frac{d}{dx}(2x-1) \right ]-\left[\frac{d}{dx}\sqrt{x^2+4} \right ](2x-1)}{x^2+4}$

$=\frac{2\sqrt{x^2+4}-\frac{1}{2\sqrt{x^2+4}}(2x)(2x-1)}{x^2+4}$

$=\frac{2\sqrt{x^2+4}-\frac{2x^2-x}{\sqrt{x^2+4}}}{x^2+4}$

$=\frac{2(x^2+4)-(2x^2-x)}{\sqrt{x^2+4}(x^2+4)}$

$=\frac{x+8}{(x^2+4)^{3/2}}$

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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 #228 : Computation Of The Derivative

Example Question #229 : Computation Of The Derivative

Example Question #230 : Computation Of The Derivative

Example Question #21 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #22 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #23 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #31 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #32 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #33 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #34 : Derivative Rules For Sums, Products, And Quotients Of Functions

All AP Calculus AB Resources

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