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

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Example Questions

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function,

$y =x^2\sin(3x)$

Possible Answers:

$\frac{dy}{dx} =3x^2\cos(3x)+2x\sin(3x)$

$\frac{dy}{dx} =-3x^2\cos(x)+2x\sin(x)$

$\frac{dy}{dx} =x^2\cos(3x)+2x\cos(3x)$

$\frac{dy}{dx} =x[3x\cos(3x)+2]$

$\frac{dy}{dx} =6x\cos(3x)$

Correct answer:

$\frac{dy}{dx} =3x^2\cos(3x)+2x\sin(3x)$

Explanation:

$y =x^2\sin(3x)$

Differentiate both sides and proceed with the product rule:

$\frac{dy}{dx}=\frac{d}{dx}[x^2\sin(3x)]$ (1)

$\frac{dy}{dx}= x^2\frac{d}{dx}\sin(3x)+ \sin(3x)\frac{d}{dx}x^2$

Evaluate the derivatives in each term. For the first term,

$=x^2\underbrace{\frac{d}{dx}\sin(3x)} + \sin(3x)\frac{d}{dx}x^2$ (2)

apply the chain rule,

$\frac{d}{dx}sin(3x)=3\cos(3x)$

So now the first term in equation (2) can be written,

$x^2\frac{d}{dx}\sin(3x)=x^2[3\cos(3x)]=3x^2\cos(3x)$ (3)

The second term in equation (2) is easy, this is just the product of $\sin(3x)$ multiplied by the derivative of x^2 ,

$\sin(3x)\frac{d}{dx}x^2 = 2x\sin(3x)$ (4)

Combine equations (3) and (4) to write the derivative,

$\frac{dy}{dx} =3x^2\cos(3x)+2x\sin(3x)$

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Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative.

$\sin (x)x^2$

Possible Answers:

$\cos (x)x+2x\sin (x)$

$\cos (x)x^2+2x^2\sin (x)$

$\cos (x)x^2+2x\sin (x)$

$\sin (x)x^2+2x\sin (x)$

Correct answer:

$\cos (x)x^2+2x\sin (x)$

Explanation:

Use the product rule to find the derivative.

$\cos (x)x^2+2x\sin (x)$

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Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative.

x^3+4x^2-x

Possible Answers:

x^2+4x-1

3x^2+8x-1

4x^4

x^2+4x

Correct answer:

3x^2+8x-1

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}x^3=3x^2$

$\frac{d}{dx}4x^2=8x$

$\frac{d}{dx}-x=-1$

Thus, the derivative is 3x^2+8x-1

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Example Question #3 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find f'(x) given $f(x)=3sin(x^2)*e^{2x-1}$

Possible Answers:

$f'(x)=6e^{2x-1}(cos(2x)+sin(x^2))$

$f'(x)=6xcos(x^2)*e^{2x-1}+3sin(x^2)*(2x-1)e^{2x-2}$

$f'(x)=12xe^{2x-1}(cos(x^2))$

$f'(x)=6e^{2x-1}(xcos(x^2)+sin(x^2))$

$f'(x)=\frac{6e^{2x-1}(xcos(x^2)*sin(x^2))}{e^{4x-2}}$

Correct answer:

$f'(x)=6e^{2x-1}(xcos(x^2)+sin(x^2))$

Explanation:

Here we use the product rule: $\frac{\mathrm{d} }{\mathrm{d} x}(F(x)\cdot G(x))=F'(x)\cdot G(x)+F(x)\cdot G'(x)$

Let F(x)=3sin(x^2) and $G(x)=e^{2x-1}$

Then $F'(x)=3cos(x^2)\cdot2x$ (using the chain rule)

and $G'(x)=e^{2x-1}\cdot2$ (using the chain rule)

Subbing these values back into our equation gives us

$f'(x)=3cos(x^2)\cdot2x\cdot e^{2x-1}+3sin(x^2)\cdot e^{2x-1}\cdot2$

Simplify by combining like-terms

$f'(x)=6xcos(x^2)\cdot e^{2x-1}+6sin(x^2)\cdot e^{2x-1}$

and pulling out a $6e^{2x-1}$ from each term gives our final answer

$f'(x)=6e^{2x-1}(xcos(x^2)+sin(x^2))$

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Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

If $f(x)=\pi\sin(ex)$ , evaluate f'(0) .

Possible Answers:

$\pi$

$e\pi$

Correct answer:

$e\pi$

Explanation:

When evaluating the derivative, pay attention to the fact that $e,\pi$ are constants, (not variables) and are treated as such.

$f'(x)=\pi\cos(ex)\times (ex)'=e\pi\cos(ex)$ .

and hence

$f'(0)=e\pi\cos(e(0))=e\pi$ .

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Example Question #2 : Derivative Rules For Sums, Products, And Quotients Of Functions

If f(x)=1+x+x^2+x^3+x^4 , evaluate f'(2)

Possible Answers:

Correct answer:

Explanation:

To obtain an expression for f'(x) , we can take the derivative of f(x) using the sum rule.

f'(x)=0+1+2x+3x^2+4x^3 .

Substituting into this equation gives us

f'(2) =0+1+4+12+32

=49 .

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Example Question #6 : Derivative Rules For Sums, Products, And Quotients Of Functions

If $f(x)=\frac{x\tan^{-1}(x)}{\ln(x)}$ , find f'(x) .

Possible Answers:

$\frac{\ln(x)(\frac{1}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{\ln(x^2)}$

$\frac{\ln(x)(\frac{1}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$

$\frac{\frac{1}{x}(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{\ln(x^2)}$

$\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$

$\frac{\ln(x)(\frac{x+1}{1+x^2})-\tan^{-1}x}{\ln(x^2)}$

Correct answer:

$\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$

Explanation:

To find f'(x) , we will need to use the quotient rule; $\frac{gf'-fg'}{g^2}$ .

$f(x)= \frac{x\tan^{-1}x}{\ln(x)}$ . Start

$f'(x)= \frac{(\ln(x))(x\tan^{-1}x)'-(x\tan^{-1}x)(\ln(x))'}{(\ln(x))^2}$ . Use the quotient rule.

$= \frac{(\ln(x))(x(\frac{1}{1+x^2})+1(\tan^{-1}x))-(x\tan^{-1}x)(\frac{1}{x})}{(\ln(x))^2}$ . Take the derivatives inside of the quotient rule. The derivative of $x\tan^{-1}x$ uses the product rule.

$=\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$ . Simplify to match the correct answer.

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Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following equation:

$h(x) = \sin(x)\cos(x)$

Possible Answers:

$\cos ^{2}(x)+\sin^{2}(x)$

$-\sin(x)\cos(x)$

$\cos ^{2}(x)$

$\cos^{2}(x)-\sin^{2}(x)$

None of the other answers

Correct answer:

$\cos^{2}(x)-\sin^{2}(x)$

Explanation:

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that $\frac{d}{dx}(f(x)g(x)) = f{}'(x)g(x) + f(x)g{}'(x)$ .

By using this rule, we get the answer:

$\cos(x)\cos(x) + \sin(x)(-\sin (x))$

By simplifying, we conclude that the derivative is equal to

$\cos ^{2}(x)-\sin ^{2}(x)$ .

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Example Question #8 : Derivative Rules For Sums, Products, And Quotients Of Functions

$\begin{align*}&\text{Calculate the derivative of the function}\\&-\frac{(114cos(19x))}{x^{20}}\end{align*}$

Possible Answers:

$-\frac{ (43320cos(19x))}{x^{21}}-\frac{ (43320sin(19x))}{x^{20}}$

$\frac{120}{x^{21}}- 361sin(19x)$

$\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}$

$-\frac{(43320sin(19x))}{x^{21}}$

Correct answer:

$\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }cos(19x)\\&\text{and the second being: }\frac{1}{x^{20}}\\&\text{With the product scaled by }-114\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[cos(ax)]=-asin(ax)dx\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=cos(19x);du=-19sin(19x)\\&v=\frac{1}{x^{20}};dv=-\frac{20}{x^{21}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}\end{align*}$

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Example Question #9 : Derivative Rules For Sums, Products, And Quotients Of Functions

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-\frac{(380tan(17x))}{2^{(20x)}}\end{align*}$

Possible Answers:

$\frac{(380ln(2))}{2^{(20x)}}+ 340tan(17x)^{2} + 340$

$\frac{(7600\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}+\frac{ (129200tan(17x)ln(2))}{2^{(20x)}}$

$\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}$

$\frac{(20ln(2)\cdot (6460tan(17x)^{2} + 6460))}{2^{(20x)}}$

Correct answer:

$\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }tan(17x)\\&\text{and the second being: }\frac{1}{2^{(20x)}}\\&\text{With the product scaled by }-380\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&\text{From the above we can continue:}\\&u=tan(17x);du=17tan(17x)^{2} + 17\\&v=\frac{1}{2^{(20x)}};dv=-\frac{(20ln(2))}{2^{(20x)}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}\end{align*}$

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

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #3 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #2 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #6 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #8 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #9 : Derivative Rules For Sums, Products, And Quotients Of Functions

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