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AP Calculus BC : Derivatives

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Example Questions

Example Question #131 : Derivatives

Find the derivative of the function

$f(x)= \frac{x^2}{x\ln(x)}$

Possible Answers:

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

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

$\frac{\ln(x)-1}{[\ln(x)]^2}$

$\frac{1-\ln(x)}{[\ln(x)]^2}$

None of the other answers.

Correct answer:

$\frac{\ln(x)-1}{[\ln(x)]^2}$

Explanation:

We find the answer using the quotient rule

$(\frac{f(x)}{g(x)})' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

and the product rule

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

and then simplifying.

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

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

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

$=\frac{\ln(x)-1}{\ln(x)^2}$ or $\frac{\ln(x)-1}{[\ln(x)]^2}$ . The extra brackets in the denominator are optional.

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Example Question #31 : First And Second Derivatives Of Functions

If x^y = xy , find in terms of and .

Possible Answers:

$\frac{xy\ln(y)^2}{\ln(x)-xy\ln(x)^2}$

$\frac{xy^2\ln(y)^2}{x\ln(x)-x^2\ln(x)^2}$

None of the other answers

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

$\frac{xy\ln(y)^2}{\ln(x)-x^2y\ln(x)^2}$

Correct answer:

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

Explanation:

Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

x^y=yx

$\ln(x^y)=\ln(xy)$

$y\ln(x)=\ln(x)+\ln(y)$

$y = \frac{\ln(y)}{\ln(x)}+1$

$y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}$ . Quotient Rule + implicit differentiation.

$y' = \frac{y'}{y\ln(x)}-\frac{\ln(y)}{x\ln(x)^2}$

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

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

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

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

$y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

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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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AP Calculus BC : Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #131 : Derivatives

Example Question #31 : First And Second Derivatives Of Functions

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

All AP Calculus BC Resources

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