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Calculus 1 : How to find differential functions

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Example Question #21 : Other Differential Functions

Find $\small d(\tan x)/dx$

Possible Answers:

$\small -\sec^2x$

$\small \csc^2x$

$\small -\csc^2x$

$\small \cot^2x$

$\small \sec^2x$

Correct answer:

$\small \sec^2x$

Explanation:

$\small \frac{d}{dx}(\tan x) = \frac{d}{dx}(\frac{\sin x}{\cos x}) = \frac {\cos x \frac{d}{dx}(\sin x) - \sin x \frac{d}{dx}(\cos x)}{\cos^2x}$

$\small = \frac{ \cos x \cos x- \sin x (- \sin x)} {\cos^2x}$

$\small = \frac{\cos^2x + \sin^2x}{\cos^2x}$
$\small = \frac{1}{\cos^2x} = \sec^2x$

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Example Question #201 : Differential Functions

Differentiate:

$\small \sin(x^2+x)$ with respect to $\small x$ .

Possible Answers:

$\small (x^2+x)(\cos(2x+1))$

$\small (2x+1)(\cos(x^2+x))$

$\small (x^2+x)(\sin(2x+1))$

$\small (2x+1)(\sin(x^2+x))$

$\small -(2x+1)(cos(x^2+x))$

Correct answer:

$\small (2x+1)(\cos(x^2+x))$

Explanation:

Apply the chain rule: differentiate the "outside" function first. Let $u=x^{2}+x$ .
$\small \small \small \small \frac{d}{du} \sin(u) = \cos(u) = \cos(x^2+x)$
Differentiate the "inside" function next.
$\small \frac{d}{dx}(x^2+x) = 2x+1$
Multiply these two functions to find the derivative of the original function.
$\small \cos(x^2+x) * (2x+1)$

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Example Question #1239 : Calculus

Evaluate $\small \int (x^2 - 3x + 6) dx$

Possible Answers:

$\small \small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x$

$\small 2x-3+C$

$\small \small \small x^3 - 3x^2 + 6x + C$

$\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

$\small 2x - 3$

Correct answer:

$\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

Explanation:

To integrate the function, integrate each term of the function. e.g., integrate $\small x^2$ by increasing the exponent by 1 integer and dividing the term by this new integer: $\small \frac {x^3}{3}$ .

Do this for the rest to get $\small \small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x$ .

But remember that every integration requires an arbitrary constant, $\small C$ . Thus, the integral of the function is $\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

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Example Question #1240 : Calculus

Integrate

$\small \int\sin^2x\hspace2dx$

Possible Answers:

$\small \small \small \frac{x}{2}-\frac{\cos2x}{4}+C$

$\small \small \small \frac{x}{2}+\frac{\cos2x}{4}+C$

$\small \small \frac{x}{2}+\frac{\sin2x}{4}+C$

$\small \frac{x}{2}-\frac{\sin2x}{4}+C$

$\small \small \small \frac{x}{2}+\frac{\sin2x}{4}$

Correct answer:

$\small \frac{x}{2}-\frac{\sin2x}{4}+C$

Explanation:

We can use trigonometric identities to transform integrals that we typically don't know how to integrate.

$\small \small \sin^2x = \frac{1-\cos2x}{2}$
Thus,
$\small \small \small \int\sin^2x\hspace2dx = \int \frac{1-\cos2x}{2}\hspace2dx$
$\small = \frac{1}{2} \int (1-\cos2x) \hspace2 dx = \frac{1}{2}x-\frac{1}{2}\frac{sin2x}{2}+C = \frac{x}{2} - \frac{sin2x}{4}+C$

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Example Question #21 : How To Find Differential Functions

Integrate

$\small \int \cos^2x \hspace2 dx$

Possible Answers:

$\small \frac{x}{2}+\frac{sin2x}{4}+C$

$\small \small \frac{x}{2}+\frac{sin2x}{4}$

$\small \small \frac{x}{2}-\frac{cos2x}{4}+C$

$\small \small \frac{x}{2}+\frac{cos2x}{4}+C$

$\small \small \frac{x}{2}-\frac{sin2x}{4}+C$

Correct answer:

$\small \frac{x}{2}+\frac{sin2x}{4}+C$

Explanation:

We can use trigonometric identities to integrate functions we typically don't know how to integrate.

$\small \cos^2x = \frac{1+\cos2x}{2}$

Thus,

$\small \small \int \cos^2x \hspace2 dx = \int \frac{1+\cos2x}{2}\hspace2dx = \frac{x}{2}+\frac{\sin2x}{4}+C$

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Example Question #22 : How To Find Differential Functions

Evaluate

$\small \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot \theta \csc^2 \theta \hspace2 d\theta$

Possible Answers:

$\small -\frac{\pi}{2}$

$\small -\frac{1}{2}$

$\small \frac{\pi}{4}$

$\small \frac{1}{2}$

Correct answer:

$\small \frac{1}{2}$

Explanation:

You can transform the limits of integration via u-substitution.

Let $\small \small u = \cot \theta, du = -\csc^2\theta\hspace2d\theta$
$\small -du = \csc^2\theta\hspace2d\theta$

When $\small \theta = \frac{\pi }{4}, u = \cot\left ( \frac{\pi }{4}\right )=1$
When $\small \theta = \frac{\pi }{2}, u = \cot\left ( \frac{\pi }{2}\right ) = 0$

Thus,

$\small \small \int_{\pi/4}^{\pi/2} \cot \theta \csc^2 \theta \hspace2 d\theta = \int_{1}^{0}u\cdot(-du)$

$\small \small = -\int_{1}^{0}u\hspace2 du$
$\small = -\left [ \frac{u^2}{2}\right ]^0_1$
$\small = \left [ \frac{(0)^2}{2} - \frac{(1)^2}{2}\right ] = \frac{1}{2}$

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Example Question #1241 : Calculus

Differentiate the function $y = x^{2}e^x$

Possible Answers:

$y^{'}=2xe^x$

$y^{'}=xe^x$

$y^{'}=2 x^2e^x$

$y^{'}=2 xe^x+x^2e^x$

$y^{'}=2e^x$

Correct answer:

$y^{'}=2 xe^x+x^2e^x$

Explanation:

Using the product rule for finding derivatives gives the answer. $f^{'}(x)g(x)+g^{'}(x)f(x)$

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Example Question #24 : How To Find Differential Functions

Solve for f{}'(x) when $f(x) = \frac{2x^2-4}{x+3}$

Possible Answers:

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

$f{}'(x)=\frac{4x^2+8x-12}{(x+3)^2}$

$f{}'(x)= \frac{4x}{4}$

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

Correct answer:

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

Explanation:

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

using the quotient rule: $(\frac{f}{g}){}'= \frac{f{}'g-fg{}'}{g^2}$

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

Foil

$f{}'(x)=\frac{4x^2+12x-2x^2+4}{(x+3)^2}$

combine like-terms to simplify

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

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Example Question #25 : How To Find Differential Functions

Solve for f{}'(x) when f(x)= 3tan(x)cot(4x)

Possible Answers:

$f{}'(x)= 3sec^2(x)cot(4x)-12tan(x)csc^2(4x)$

$f{}'(x)= 3sec^2(x)cot(4x)-3tan(x)csc^2(4x)$

$f{}'(x)= 3sec^2(x)cot(4x)+12tan(x)csc^2(4x)$

$f{}'(x)= sec^2(x)cot(4x)-4tan(x)csc^2(4x)$

Correct answer:

$f{}'(x)= 3sec^2(x)cot(4x)-12tan(x)csc^2(4x)$

Explanation:

f(x)= 3tan(x)cot(4x)

Using the Product Rule: (fg){}'=f{}'g+fg{}' and chain rule for trignometry functions: $\frac{d}{dx}(cot[f(x)])=f{}'(x)cot[f(x)]$

$f{}'(x)=(3sec^2(x))(cot(4x))+(3tan(x))(4)(-csc^2(4x))$

Simplify

$f{}'(x)= 3sec^2(x)cot(4x)-12tan(x)csc^2(4x)$

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Example Question #26 : How To Find Differential Functions

Find the derivative of $y = sin((\sqrt{x})^3)$

Possible Answers:

$y^{'}=\frac{\sqrt{x}\cdot sin(x^\frac{3}{2})}{2}$

$y^{'}=\frac{2\sqrt{x}\cdot sin(x^\frac{3}{2})}{3}$

$y^{'}=\frac{3\sqrt{x}\cdot cos(x^\frac{3}{2})}{2}$

$y^{'}=\frac{3\sqrt{x}\cdot sin(x)}{2}$

$y^{'}=\frac{3cos(x^\frac{3}{2})}{\sqrt{x}}$

Correct answer:

$y^{'}=\frac{3\sqrt{x}\cdot cos(x^\frac{3}{2})}{2}$

Explanation:

The quantity square root of raised to the third is the same as $x^{\frac{3}{2}}$ . Using the chain rule and power rule, the answer can be found.

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : Other Differential Functions

Example Question #201 : Differential Functions

Example Question #1239 : Calculus

Example Question #1240 : Calculus

Example Question #21 : How To Find Differential Functions

Example Question #22 : How To Find Differential Functions

Example Question #1241 : Calculus

Example Question #24 : How To Find Differential Functions

Example Question #25 : How To Find Differential Functions

Example Question #26 : How To Find Differential Functions

All Calculus 1 Resources

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