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Calculus AB : Calculus AB

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Example Questions

Example Question #31 : Calculus Ab

Find the slope of the tangent line to the function $h(x)=\frac{2ln(x)}{x^2}$ at x=1 .

Possible Answers:

ln(2)-1

Correct answer:

Explanation:

The slope of the tangent line to a function at a point is the value of the derivative of the function at that point. In this problem, h(x) is a quotient of two functions, $2ln(x) \ and\ x^2$ , so the quotient rule is needed.

In general, the quotient rule is

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$ .

To apply the quotient rule in this example, you must also know that $\frac{d}{dx}ln(x)=\frac{1}{x}$ and that $\frac{d}{dx}x^n=nx^{n-1}$ .

Therefore, the derivative is

$h'(x)=\frac{x^2(\frac{2}{x})-2ln(x)(2x)}{(x^2)^2}=\frac{2x-4xln(x)}{x^4}$

The last step is to substitute for in the derivative, which will tell us the slope of the tangent line to h(x) at x = 1 .

$h'(1)=\frac{2(1)-4(1)ln(1)}{1^4}=\frac{2-0}{1}=2$

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Example Question #32 : Calculus Ab

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

Possible Answers:

e^x (4x-1)

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

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

e^x (2x^2 +3x-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 #33 : Calculus Ab

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

Possible Answers:

$2 \cos (2x )$

$4 \cos ^2 x$

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 #34 : Calculus Ab

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

Possible Answers:

$\frac{1}{4}$

$\frac{1}{e}$

$\frac{1}{2}$

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 #35 : Calculus Ab

Find the derivative of f:

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

Possible Answers:

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

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

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

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

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 #36 : Calculus Ab

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 #37 : Calculus Ab

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

1+2sin(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 #38 : Calculus Ab

Suppose and are differentiable functions, and f(3)=1, f'(3)=4, g(-1)=3, g'(-1)=-1 .

Calculate the derivative of h(x) = f(x+2)g(-x) , at x = 1 .

Possible Answers:

None of the other answers

Correct answer:

None of the other answers

Explanation:

The correct answer is 11.

Taking the derivative of h(x) involves the product rule, and the chain rule.

h'(x)=[f(x+2)][-g'(-x)]+[g(-x)][f'(x+2)]

= -g'(-x)f(x+2)+g(-x)f'(x+2)

Substituting x=1 into both sides of the derivative we get

$h'(1)= -g'(-1)f(3)+g(-1)f'(3) = -1\cdot1 + 3 \cdot 4 = 11$

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Example Question #39 : Calculus Ab

Find the second derivative of g(x)

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

Possible Answers:

36xln(x)+21x^2

36x^2ln(x)+21x

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

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

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 #40 : Calculus Ab

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

Possible Answers:

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

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

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

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

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

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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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #31 : Calculus Ab

Example Question #32 : Calculus Ab

Example Question #33 : Calculus Ab

Example Question #34 : Calculus Ab

Example Question #35 : Calculus Ab

Example Question #36 : Calculus Ab

Example Question #37 : Calculus Ab

Example Question #38 : Calculus Ab

Example Question #39 : Calculus Ab

Example Question #40 : Calculus Ab

All Calculus AB Resources

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