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AP Calculus BC : Computation of Derivatives

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Example Questions

Example Question #32 : First And Second Derivatives Of Functions

Find the derivative of the function $y= \frac{\tan^{-1}(x)}{x}$

Possible Answers:

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

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

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

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

None of the other answers

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}$ .

Using the Quotient Rule and the fact $\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$ , we have

$y = \frac{\tan^{-1}(x)}{x}$

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

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

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

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

$y' = \frac{x-\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}$ .

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Example Question #401 : Ap Calculus Bc

$\frac{\mathrm{d} }{\mathrm{d} x} \enspace 2 e^x x^2 -xe^x$

Possible Answers:

e^x (2x^2 -x) + 4x - 1

e^x (2x^2 +3x-1)

e^x (4x-1)

e^x(3x^2 + 2x - 1 )

Correct answer:

e^x (2x^2 +3x-1)

Explanation:

First, factor out e^x : e^x (2x^2 -x ) . Now we can differentiate using the product rule, f(x)g(x) '= f'(x)g(x) + f(x)g'(x) .

Here, f(x) = e^x so f'(x) = e^x . g(x) = 2x^2 - x so g'(x) = 4x - 1 .

The answer is e^x (2x^2 -x) + e^x (4x - 1 ) = e^x [ 2x^2 - x + 4x - 1 ] = e^x (2x^2 + 3x -1)

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Example Question #402 : Ap Calculus Bc

$\frac{\mathrm{d} }{\mathrm{d} x} 2 \sin x \cos x$

Possible Answers:

$4 \cos ^2 x$

$2 \cos (2x )$

Correct answer:

$2 \cos (2x )$

Explanation:

According to the product rule, f(x)g(x) ' = f'(x)g(x) + f(x)g'(x) . Here $f(x) = 2 \sin x$ so $f'(x) = 2 \cos x$ . $g(x) = \cos x$ so $g'(x) = -\sin x$ .

The derivative is $2 \cos x \cdot \cos x + 2 \sin x \cdot - \sin x = 2 \cos ^2 x - 2 \sin ^2 x$

Factoring out the 2 gives $2(\cos ^2 x - \sin ^2 x )$ . Remembering the double angle trigonometric identity finally gives $2 (\cos (2x ))$

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Example Question #403 : Ap Calculus Bc

If $f(t) = \frac{t}{\ln(t)}$ , find f'(e^2)

Possible Answers:

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{e}$

Correct answer:

$\frac{1}{4}$

Explanation:

First, we need to find f'(t) . We can do that by using the quotient rule.

$f'(t) = \frac{(\ln(t))(1)-(t)(\frac{1}{t})}{(\ln(t))^2}$ .

Plugging e^2 in for and simplifying, we get

$f'(e^2) = \frac{(\ln(e^2))(1)-(e^2)(\frac{1}{e^2})}{(\ln(e^2))^2} = \frac{2-1}{2^2} = \frac{1}{4}$ .

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Example Question #404 : Ap Calculus Bc

Find the derivative of f:

$f=\frac{x^3}{x\cos(x)}$

Possible Answers:

$\frac{3x^3\cos(x)-x^4\sin(x)}{x^2\cos^2(x)}$

$\frac{2x^3\cos(x)+x^4\sin(x)}{x\cos(x)}$

$\frac{2x^3\cos(x)+x^4\sin(x)}{x^2\cos^2(x)}$

$\frac{x^3\cos(x)+x^4\sin(x)}{x^2\cos^2(x)}$

Correct answer:

$\frac{2x^3\cos(x)+x^4\sin(x)}{x^2\cos^2(x)}$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #405 : Ap Calculus Bc

Find the derivative of the function:

$f=x^3+x^2+\alpha^2+2\alpha^2x$

where $\alpha$ is a constant

Possible Answers:

3x^2+2x

$3x^2+2x+\alpha+2\alpha^2$

$3x^2+2x+2\alpha^2$

$x^2+2x+2\alpha^2$

Correct answer:

$3x^2+2x+2\alpha^2$

Explanation:

When taking the derivative of the sum, we simply take the derivative of each component.

The derivative of the function is

$f'=3x^2+2x+2\alpha^2$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #406 : Ap Calculus Bc

Compute the first derivative of the following function.

f(x)=sin(x)cox(x)+x^2sin(x)

Possible Answers:

cos^2(x)-sin^2(x)+2xsin(x)-x^2cos(x)

cos^2(x)-sin^2(x)+2xsin(x)+x^2cos(x)

1+2sin(x)+x^2cos(x)

1+2xsin(x)+x^2cos(x)

Correct answer:

cos^2(x)-sin^2(x)+2xsin(x)+x^2cos(x)

Explanation:

Compute the first derivative of the following function.

f(x)=sin(x)cox(x)+x^2sin(x)

To solve this problem, we need to apply the product rule:

(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)

So, we need to apply this rule to each of the terms in our function. Let's start with the first term

(sin(x)cos(x))'=cos(x)cos(x)+sin(x)(-sin(x))=cos^2(x)-sin^2(x)

Next, let's tackle the second part

(x^2sin(x))=2xsin(x)+x^2cos(x)

Now, combine the two to get:

cos^2(x)-sin^2(x)+2xsin(x)+x^2cos(x)

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Example Question #407 : Ap Calculus Bc

Evaluate the derivative of the function $y = x^2\cos(3x)$ .

Possible Answers:

$x^2\cos(3x)-2x\sin(3x)$

$2x\cos(3x)+3x^2\sin(3x)$

$2x\cos(3x)-3x^2\sin(3x)$

$2x\cos(3x)-x^2\sin(3x)$

$2x\cos(3x)+x^2\sin(3x)$

Correct answer:

$2x\cos(3x)-3x^2\sin(3x)$

Explanation:

Use the product rule: (fg)' = f'g + fg'

where f(x) = x^2 and $g(x) = \cos(3x)$ .

By the power rule, f'(x)=2x .

By the chain rule, $g'(x) = -\sin(3x)\cdot3$ .

Therefore, the derivative of the entire function is:

$y'=(2x)(\cos(3x))+(x^2)(-\sin(3x)\cdot 3)$

$y'=2x\cos(3x)-3x^2\sin(3x)$ .

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Example Question #408 : Ap Calculus Bc

Find the second derivative of g(x)

g(x)=3x^4ln(x)

Possible Answers:

12x^3ln(x)+3x^3

36x^2ln(x)+21x

36x^2ln(x)+21x^2

36xln(x)+21x^2

Correct answer:

36x^2ln(x)+21x^2

Explanation:

Find the second derivative of g(x)

g(x)=3x^4ln(x)

To find this derivative, we need to use the product rule:

(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)

So, let's begin:

g(x)=3x^4ln(x)

$g'(x)=12x^3ln(x)+3x^4(\frac{1}{x})=12x^3ln(x)+3x^3$

So, we are closer, but we need to derive again to get the 2nd derivative

g'(x)=12x^3ln(x)+3x^3

$g''(x)=36x^2ln(x)+12x^3(\frac{1}{x})+9x^2=36x^2ln(x)+21x^2$

So, our answer is:

36x^2ln(x)+21x^2

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Example Question #12 : Derivatives

Find dy/dx by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Possible Answers:

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{e^y+3x^2}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{e^y+\tan^{-1}(y)-x^3}{x}$

$\frac{3x^2-\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Correct answer:

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Explanation:

To find dy/dx we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dx}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dx}))+(1)(\frac{dx}{dx})(\tan^{-1}(y))]=e^y\frac{dy}{dx}$

Notice that anytime we take the derivative of a term with x involved we place a "dx/dx" next to it, but this is equal to "1".

So this now becomes:
$3x^2+[\frac{x}{1+y^2}\frac{dy}{dx}+\tan^{-1}(y)}]=e^y\frac{dy}{dx}$

Now if we place all the terms with a "dy/dx" onto one side and factor out we can solved for it:

$3x^2+\tan^{-1}(y)=e^y\frac{dy}{dx}-\frac{x}{1+y^2}\frac{dy}{dx}$

$3x^2+\tan^{-1}(y)=\frac{dy}{dx}(e^y-\frac{x}{1+y^2})$

$\frac{dy}{dx}=\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

This is one of the answer choices.

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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 #32 : First And Second Derivatives Of Functions

Example Question #401 : Ap Calculus Bc

Example Question #402 : Ap Calculus Bc

Example Question #403 : Ap Calculus Bc

Example Question #404 : Ap Calculus Bc

Example Question #405 : Ap Calculus Bc

Example Question #406 : Ap Calculus Bc

Example Question #407 : Ap Calculus Bc

Example Question #408 : Ap Calculus Bc

Example Question #12 : Derivatives

Find dy/dx by implicit differentiation:

All AP Calculus BC Resources

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