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Calculus 1 : Differential Functions

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Example Questions

Example Question #211 : Differential Functions

Integrate

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

Possible Answers:

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

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

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

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

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

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 #25 : Other Differential Functions

Integrate

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

Possible Answers:

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

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

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

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

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

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 #212 : Differential Functions

Evaluate

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

Possible Answers:

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

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

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

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

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 #211 : Differential Functions

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

Possible Answers:

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

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

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

$y^{'}=xe^x$

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

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 #212 : Functions

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

Possible Answers:

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

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

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

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

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 #213 : Functions

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

Possible Answers:

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

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

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

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

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 #216 : Differential Functions

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

Possible Answers:

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

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

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

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

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

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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Example Question #214 : Functions

Solve for f{}'(x) when

$f(x)= \left(9x^4+4x^3+\frac{x}{2}\right)^9$

Possible Answers:

$f{}'(x)=36x^3+12x^2+\frac{1}{2}$

$f{}'(x)=9(9x^4+4x^3+\frac{x}{2})^8$

$f{}'(x)=(9x^4+4x^3+\frac{x}{2})^8(324x^3+108x^2+\frac{9}{2})$

$f{}'(x)=(9x^4+4x^3+\frac{x}{2})^9(36x^3+12x^2+\frac{1}{2})$

Correct answer:

$f{}'(x)=(9x^4+4x^3+\frac{x}{2})^8(324x^3+108x^2+\frac{9}{2})$

Explanation:

$f(x)= \left(9x^4+4x^3+\frac{x}{2}\right)^9$

using the chain rule:

$\frac{d}{dx}(f(g(x)))=f{}'(g(x))g{}'(x)$

$f{}'(x)=9(9x^4+4x^3+x/2)^8(36x^3+12x^2+1/2)$

multiply the constant 9 into the second function to simplify answer

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Example Question #31 : How To Find Differential Functions

Solve for f{}'(x) when $f(x)=\log_{9}x^2$

Possible Answers:

$f{}'(x)=\frac{2}{x\ln 9}$

$f{}'(x)=\frac{2}{\ln 9}$

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

$f{}'(x)=\frac{x}{\ln 9}$

Correct answer:

$f{}'(x)=\frac{2}{x\ln 9}$

Explanation:

$f(x)=\log_{9}x^2$

using the logarithm identities change the equation to base 10:

$f(x)=\frac{\ln x^2}{\ln 9}$

using lograthim identities simplify the numerator:

$f(x)=\frac{2\ln x}{\ln 9}$

differentiate

$f{}'(x)=\frac{2*\frac{1}{x}}{\ln 9}$

Simplify

$f{}'(x)=\frac{2}{x\ln 9}$

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Example Question #33 : Other Differential Functions

Solve for when

$f(x)= (x^2 +7x)(3x^3+\frac{x}{4})$

Possible Answers:

$f^{'}(x)=15x^{4}+84x^{3}+\frac{3x^{2}}{4}+\frac{7x}{2}$

$f{}'(x)= 15x^4 + 21x^3 +255x^2 + 7x +7$

$f'(x) = 81x^2 + \frac{x}{2}+\frac{7}{4}$

$f{}'(x)= 6x^4 +21x^3 +\frac{x^2}{2}+\frac{7x}{4}$

Correct answer:

$f^{'}(x)=15x^{4}+84x^{3}+\frac{3x^{2}}{4}+\frac{7x}{2}$

Explanation:

$f(x)= (x^2 +7x)(3x^3+\frac{x}{4})$

Using the product rule: (fg){}'=f{}'g+fg{}'

$f{}'(x)= (2x+7)(3x^3 + x/4) + (x^2+7x)(9x^2+1/4)$

FOIL

$f{}'(x)= 6x^4+2x^2/4+21x^3+7x/4+9x^4+x^2/4+63x^3+7x/4$

Combine like-terms

$f{}'(x)= 15x^4 + 84x^3+\frac{3x^2}{4}+\frac{14x}{4}$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #211 : Differential Functions

Example Question #25 : Other Differential Functions

Example Question #212 : Differential Functions

Example Question #211 : Differential Functions

Example Question #212 : Functions

Example Question #213 : Functions

Example Question #216 : Differential Functions

Example Question #214 : Functions

Example Question #31 : How To Find Differential Functions

Example Question #33 : Other Differential Functions

Solve for when

$f(x)= (x^2 +7x)(3x^3+\frac{x}{4})$

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