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Calculus 1 : Functions

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Example Questions

Example Question #241 : Functions

Solve the indefinite integral. If you cannot evaluate directly, use u-substitution.

$\int \frac{ln(x)}{x}dx$

Possible Answers:

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

$\frac{1}{ln(x)} +c$

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

ln(ln(x))+c

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

Correct answer:

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

Explanation:

$\int \frac{ln(x)}{x}dx$

$Let \ u = ln(x), \ du=\frac{1}{x}dx$ .

$\int \frac{ln(x)}{x}dx = \int udu = \frac{1}{2}u^2 + c = \frac{1}{2}ln(x)^2 +c$ , which is our final answer.

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Example Question #1271 : Calculus

Solve the indefinite integral. If you cannot evaluate directly, use u-substitution.

$\int \frac{e^{x^{\frac{1}{2}}}}{x^{\frac{1}{2}}}dx$

Possible Answers:

$e^{x^{\frac{3}{2}}} + c$

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

$\frac{1}{2}e^{x^{\frac{1}{2}}} + c$

$2e^{x^{\frac{1}{2}}} + c$

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

Correct answer:

$2e^{x^{\frac{1}{2}}} + c$

Explanation:

$\int \frac{e^{x^{\frac{1}{2}}}}{x^{\frac{1}{2}}}dx$

We rewrite the denominator as a negative exponenet in the numerator to make the u-substitution easier to see:

$\int \frac{e^{x^{\frac{1}{2}}}}{x^{\frac{1}{2}}}dx = \int e^{x^{\frac{1}{2}}}}{x^{- \frac{1}{2}}dx$

$Let\: u=x^{1/2}, du=\frac{1}{2}x^{-1/2}dx.$

$\int e^{x^{\frac{1}{2}}}}{x^{- \frac{1}{2}}dx = 2\int \frac{1}{2}e^{x^{\frac{1}{2}}}}{x^{- \frac{1}{2}}dx = 2 \int e^udu = 2e^u + c = 2e^{x^{\frac{1}{2}}} + c$ , which is our final answer.

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Example Question #242 : Functions

Solve the indefinite integral. If you cannot evaluate directly, use u-substitution.

$\int 2sinx*cosxdx$

Possible Answers:

sin^2x +c

ln(sinx)+c

tanx +c

-2sinx*cosx +c

2sinx*cosx +c

Correct answer:

sin^2x +c

Explanation:

$\int 2sinx*cosxdx$

$Let\: u=sinx,\: du=cosxdx.$

$\int 2sinx*cosxdx = \int 2udu = u^2 +c = sin^2x +c$ , which is our final answer.

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Example Question #241 : Differential Functions

Solve the indefinite integral. If you cannot evaluate directly, use u-substitution.

$\int \frac{e^{(1+lnx)}}{x}dx$

Possible Answers:

$\frac{-e^{(1+lnx)}}{x^2}+c$

$\frac{x}{1+lnx}+c$

$e^{(1+lnx)}+c$

$e^{(x+xlnx)}+c$

1+lnx+c

Correct answer:

$e^{(1+lnx)}+c$

Explanation:

$\int \frac{e^{(1+lnx)}}{x}dx$

$Let \ u = 1+ln(x)), \ du=\frac{1}{x}dx.$

$\int \frac{e^{(1+lnx)}}{x}dx = \int e^udu= e^u +c = e^{(1+lnx)}+c$ , which is our final answer.

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Example Question #241 : Functions

Find the derivative of x^2 - sin(x) .

Possible Answers:

2x+sin(x)

2x+cos(x)

2x - cos(x)

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

2x - sin(x)

Correct answer:

2x - cos(x)

Explanation:

The derivative of the difference of two functions is the difference of the derivative of the two functions:
$\frac{dy}{dx} = 2x - \frac{d}{dx}(sin(x))$
= 2x - cos(x)

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Example Question #242 : Functions

Find the derivative of $\frac{1}{(3x^2+5)^4}$ .

Possible Answers:

$\frac{24x}{(3x^2+5)^5}$

$\frac{-24x}{(3x^2+5)^5}$

$\frac{-24x}{(3x^2+5)^4}$

$\frac{-4x}{(3x^2+5)^5}$

Correct answer:

$\frac{-24x}{(3x^2+5)^5}$

Explanation:

We can write the function as

$\frac{1}{(3x^2+5)^4}=(3x^2+5)^{-4}}$ .

Let $u(x)=3x^2+5\Rightarrow \frac{du}{dx}=6x$ .

We then have

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}=-4(3x^2+5)^{-5}*(6x)=\frac{-24x}{(3x^2+5)^5}$ .

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Example Question #52 : Other Differential Functions

Differentiate the function:

$f(x)=e^{x^{2}}$

Possible Answers:

$f'(x)= e^{2x}$

$f'(x)= 2x*e^{2x}$

$f'(x)= 2x*e^{x^{2}}$

$f'(x)= x^{2}*e^{x^{2}}$

$f'(x)= e^{x^{2}}$

Correct answer:

$f'(x)= 2x*e^{x^{2}}$

Explanation:

$f(x)=e^{x^{2}}$

Using the chain rule, $if \ f(x)=g(h(x))$ , $then\ f'(x)=g'(h(x))*h'(x)$ ,

we observe the following:

$g(x)= e^x \ so\ g'(x)= e^x$ .

$h(x) = x^2 \ so \ h'(x)=2x$ .

$f'(x)= 2x*e^{x^{2}}$

$f'(x)=g'(h(x))*h'(x) = e^{x^{2}}*2x = 2x*e^{x^{2}}$ , which is our final answer.

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Example Question #243 : Functions

Differentiate the function:

$f(x)=\frac{sin(x)}{x}$

Possible Answers:

$f'(x)=\frac{x*sin(x) - cos(x)}{x}$

$f'(x)=\frac{ sin(x) - x*cos(x)}{x^2}$

$f'(x)=\frac{ sin(x) + cos(x)}{x}$

f'(x)=1

$f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$

Correct answer:

$f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$

Explanation:

$f(x)=\frac{sin(x)}{x}$

We evaluate this derivative using the quotient rule:

$if \ f(x)=\frac{g(x)}{h(x)}$ ,

$then\ f'(x)=\frac{g'(x)*h(x) - g(x)*h'(x)}{h(x)^{2}}$ .

Apply the above formula:

$f'(x)=\frac{cos(x)*x - sin(x)*1}{x^2}$

$f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$ , which is our final answer.

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Example Question #245 : Functions

What is the slope of the line tangent to f(x) = x⁴ – 3x^–4 – 45 at x = 5?

Possible Answers:

355.096

400.096

355.00384

500.00384

422.125

Correct answer:

500.00384

Explanation:

First we must find the first derivative of f(x).

f'(x) = 4x³ + 12x^–5

To find the slope of the tangent line of f(x) at 5, we merely have to evaluate f'(x) at 5:

f'(5) = 4*5³ + 12* 5^–5 = 500 + 12/3125 = 500.00384

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Example Question #63 : Other Differential Functions

Solve for g{}'(x) when

$g(x)=tan^{-1}(4x^3+7x^2)$

Possible Answers:

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

$g{}'(x)=sec^2(4x^3+7x^2)$

$g{}'(x)=\frac{4x^3+7x^2}{(12x^2+14x)^2}$

$g{}'(x)=sec^2(12x^2+14x)$

Correct answer:

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

Explanation:

$g{}'(x)=tan^{-1}(4x^3+7x^2)$

using the identity:

$\frac{\mathrm{d} }{\mathrm{d} x}\ tan^{-1}[(f(x)]=\frac{f{}'(x)}{1+[f(x)]^2}$

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #241 : Functions

Example Question #1271 : Calculus

Example Question #242 : Functions

Example Question #241 : Differential Functions

Example Question #241 : Functions

Example Question #242 : Functions

Example Question #52 : Other Differential Functions

Example Question #243 : Functions

Example Question #245 : Functions

Example Question #63 : Other Differential Functions

All Calculus 1 Resources

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