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AP Calculus AB : Derivative rules for sums, products, and quotients of functions

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Example Question #707 : Ap Calculus Ab

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

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

Possible Answers:

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

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

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

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

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

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 #708 : Ap Calculus Ab

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

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

Possible Answers:

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

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

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

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

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

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 #401 : Derivatives

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

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

Possible Answers:

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

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

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

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

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

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 #710 : Ap Calculus Ab

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

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

Possible Answers:

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

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

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

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

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

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

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

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

Possible Answers:

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

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

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

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

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

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

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

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

Possible Answers:

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

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

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

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

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

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

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

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

Possible Answers:

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

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

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

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

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

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 #24 : 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:

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

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

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

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

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

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 #25 : 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

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

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

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

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 #231 : Computation Of The Derivative

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

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

Possible Answers:

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

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

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

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

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

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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AP Calculus AB : Derivative rules for sums, products, and quotients of functions

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #707 : Ap Calculus Ab

Example Question #708 : Ap Calculus Ab

Example Question #401 : Derivatives

Example Question #710 : Ap Calculus Ab

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

Example Question #24 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #25 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #231 : Computation Of The Derivative

All AP Calculus AB Resources

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