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AP Calculus BC : Computation of Derivatives

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

Example Question #12 : Chain Rule And Implicit Differentiation

$\frac{d}{dx } \tan ^ 3 x$

Possible Answers:

$2\tan^2 x$

$3 \tan ^2 x$

$1 + \tan ^2 x$

$3 \tan ^2 x + 3 \tan ^4 x$

Correct answer:

$3 \tan ^2 x + 3 \tan ^4 x$

Explanation:

According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, f(x) = x^3 and $g(x) = \tan x$ . The derivative is $3 (\tan x) ^2 \cdot( 1 + \tan ^2 x ) = 3 \tan ^2 x (1 + \tan ^2 x ) = 3 \tan ^2 x + 3 \tan ^4 x$

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Example Question #13 : Chain Rule And Implicit Differentiation

$\frac{d}{dx } 4(x-2)^2$

Possible Answers:

8x - 16

-16(x-2)

Correct answer:

8x - 16

Explanation:

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, f(x ) = 4x^2 and g(x) = x-2 . f'(x) = 8x and g'(x) = 1 .

The derivative is $8(x-2) \cdot 1 = 8x -16$

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Example Question #151 : Derivatives

$\frac{d}{dx} e^{\sin x}$

Possible Answers:

$e ^ {\sin x }$

$- \cos x \cdot e ^ {\sin x }$

$e ^{ \cos x }$

$\cos x \cdot e^ {\sin x}$

Correct answer:

$\cos x \cdot e^ {\sin x}$

Explanation:

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, f(x) = e^x and $g(x) = \sin x$ . Since f'(x) = e^x and $g'(x) = \cos x$ , the derivative is $e ^ {\sin x } \cdot \cos x$

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Example Question #14 : Chain Rule And Implicit Differentiation

$\frac{\mathrm{d} }{\mathrm{d} x} 3 ^ {x^2 }$

Possible Answers:

$2x \ln 3$

$3^{x^2} \cdot 2x \cdot \ln(3)$

$3^ {2x} \cdot \ln(3)$

$3^{x^2} (\frac{x^2}{3} + 2x \ln 3)$

Correct answer:

$3^{x^2} \cdot 2x \cdot \ln(3)$

Explanation:

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, f(x) = 3^x and g(x) = x^2 . Since $f'(x) = 3^ {x} \cdot \ln(3)$ and g'(x) = 2x , the derivative is $3^{x^2 } \cdot \ln(3 ) \cdot 2x$

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Example Question #15 : Chain Rule And Implicit Differentiation

$\frac{\mathrm{d} }{\mathrm{d} x} \enspace \frac{1}{4} \cdot 2^{4x-1}$

Possible Answers:

$2^ {4x-1} \cdot \ln(2)$

$\frac{\ln(2)}{4}$

$\frac{2^{4x-1}\cdot \ln (2)}{2}$

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

Correct answer:

$2^ {4x-1} \cdot \ln(2)$

Explanation:

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, $f(x) = \frac{1}{4} \cdot 2^x$ and g(x) = 4x-1 . Here $f'(x) = \frac{1}{4} \cdot 2^x \cdot \ln(2)$ and g'(x) = 4 . The derivative is:

$\frac{1}{4} \cdot 2 ^{4x-1} \cdot \ln(2) \cdot 4 = 2^{4x-1} \cdot \ln(2)$

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Example Question #16 : Chain Rule And Implicit Differentiation

Given the relation y^2=x^2+y , find $\frac{dy}{dx}$ .

Possible Answers:

$\frac{2x}{y}$

$\frac{y}{x}$

$\frac{x}{y-1}$

$\frac{2x}{2y-1}$

Correct answer:

$\frac{2x}{2y-1}$

Explanation:

We begin by taking the derivative of both sides of the equation.

$\frac{d}{dx}(y^2)=\frac{d}{dx}(x^2+y)$ .

2yy' = 2x +y' . (The left hand side uses the Chain Rule.)

2yy' -y' = 2x .

(2y-1)y'=2x

$y' = \frac{2x}{2y-1}$

$\frac{dy}{dx} = \frac{2x}{2y-1}$ .

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Example Question #17 : Chain Rule And Implicit Differentiation

Given the relation x^2+xy^2=y , find .

Possible Answers:

$\frac{2x}{2x+y}$

None of the other answers

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

$\frac{x+y}{y^2}$

$\frac{2x+1}{2x+y^2}$

Correct answer:

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

Explanation:

We can use implicit differentiation to find . We being by taking the derivative of both sides of the equation.

$\frac{d}{dx}(x^2+xy^2)=\frac{d}{dx}(y)$

$\frac{d}{dx}(x^2)+\frac{d}{dx}(xy^2)=y'$

$2x +[x \times 2y' + y^2 \times 1] = y'$ (This line uses the product rule for the derivative of xy^2 .)

2x +2xy' +y^2 = y'

2x+y^2 = -2xy' + y'

2x+y^2 = (-2x+1)y'

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

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Example Question #18 : Chain Rule And Implicit Differentiation

If $f(x) = e^{e^x}$ , find f'(1) .

Possible Answers:

$e^{e+1}$

e^e

e^2

Correct answer:

$e^{e+1}$

Explanation:

Since we have a function inside of a another function, the chain rule is appropriate here.

The chain rule formula is

$(h(g(x)))' = h'(g(x))\times g'(x)$ .

In our function, both h(x), g(x) are e^x

So we have

$f'(x) = (h(g(x)))' = e^{e^x} \times e^x = e^{e^x+1}$

and

$f'(1)=e^{e+1}$ .

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Example Question #19 : Chain Rule And Implicit Differentiation

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ of the following:

xe^y+y^2+x^2=y

Possible Answers:

$\frac{-e^y-2x}{xe^y+2y+1}$

e^y+xe^y+2y+2x

$\frac{-e^y-2x}{xe^y+2y-1}$

$\frac{-e^y-2x}{xe^y+2y}$

Correct answer:

$\frac{-e^y-2x}{xe^y+2y-1}$

Explanation:

To find $\frac{\mathrm{d} y}{\mathrm{d} x}$ we must use implicit differentiation, which is an application of the chain rule.
Taking $\frac{\mathrm{d} }{\mathrm{d} x}$ of both sides of the equation, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=e^y+xe^y\frac{\mathrm{d} y}{\mathrm{d} x}+2y\frac{\mathrm{d} y}{\mathrm{d} x}+2x$

using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that for every derivative with respect to x of a function with y, the additional term dy/dx appears; this is because of the chain rule, where y=g(x), so to speak, for the function it appears in.

Using algebra, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-e^y-2x}{xe^y+2y-1}$

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Example Question #21 : Chain Rule And Implicit Differentiation

Find :

$f=\gamma \cos(x^2)$ , where $\gamma$ is a constant.

Possible Answers:

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

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

$-2x\gamma \sin(x^2)$

$-\gamma \sin(x^2)$

$2x\gamma \cos(x^2)$

Correct answer:

$-2x\gamma \sin(x^2)$

Explanation:

The derivative of the function is equal to

$f'=-2x\gamma \sin(x^2)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The constant may seem intimidating, but we treat it as another constant!

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AP Calculus BC : Computation of Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #12 : Chain Rule And Implicit Differentiation

Example Question #13 : Chain Rule And Implicit Differentiation

Example Question #151 : Derivatives

Example Question #14 : Chain Rule And Implicit Differentiation

Example Question #15 : Chain Rule And Implicit Differentiation

Example Question #16 : Chain Rule And Implicit Differentiation

Example Question #17 : Chain Rule And Implicit Differentiation

Example Question #18 : Chain Rule And Implicit Differentiation

Example Question #19 : Chain Rule And Implicit Differentiation

Example Question #21 : Chain Rule And Implicit Differentiation

All AP Calculus BC Resources

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