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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

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 #1271 : Calculus

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

Possible Answers:

2x - cos(x)

2x - sin(x)

2x+sin(x)

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

2x+cos(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 #1272 : Calculus

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

Possible Answers:

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

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

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

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

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 #1273 : Calculus

Differentiate the function:

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

Possible Answers:

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

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

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

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

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

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 #1274 : Calculus

Differentiate the function:

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

Possible Answers:

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

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

f'(x)=1

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

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

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 #1275 : Calculus

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

Possible Answers:

422.125

355.096

400.096

500.00384

355.00384

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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Example Question #251 : Differential Functions

Differentiate $\small f(x)=\frac{x^2}{2x+7}$

Possible Answers:

$\small f'(x)=\frac{-2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{2x(x+7)}{(2x+7)}$

$\small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{2x(3x+7)}{(2x+7)^2}$

Correct answer:

$\small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

Explanation:

The Quotient Rule applies when differentiating quotients of functions. Here, $\small f(x)$ equals the quotient of two functions, $\small x^2$ and $\small 2x+7$ . Let $\small l(x)=2x+7$ and $\small h(x)=x^2$ . (Think: $\small l(x)$ is the "low" function or denominator and $\small h(x)$ is the "high" function or numerator.) The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function. In other words,

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

Here, $\small l(x)=2x+7$ so $\small l'(x)=2$ . Similarly, $\small h(x)=x^2$ so $\small h'(x)=2x$ .

Then

$\small f'(x)=\frac{(2x+7)(2x)-x^2(2)}{(2x+7)^2}$

$\small f'(x)=\frac{(4x^2+14x)-2x^2}{(2x+7)^2}$

$\small f'(x)=\frac{2x^2+14x}{(2x+7)^2}$

Factoring out $\small 2x$ from the numerator gives

$\small \small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{-2x(x+7)}{(2x+7)^2}$ inverts the order of the numerator, subtracting $\small l(x)h'(x)$ from $\small h(x)l'(x)$ .

$\small f'(x)=\frac{2x(3x+7)}{(2x+7)^2}$ adds the products in the numerator, rather than subtracting them.

$\small f'(x)=\frac{2x(x+7)}{(2x+7)}$ fails to square the denominator.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1271 : Calculus

Example Question #242 : Functions

Example Question #241 : Differential Functions

Example Question #1271 : Calculus

Example Question #1272 : Calculus

Example Question #1273 : Calculus

Example Question #1274 : Calculus

Example Question #1275 : Calculus

Example Question #63 : Other Differential Functions

Example Question #251 : Differential Functions

All Calculus 1 Resources

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