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Calculus 2 : Other Derivative Review

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Example Questions

Example Question #11 : Other Derivative Review

Let f(x)=cos^3(x+1) . Determine the value of $f'(\pi)$ .

Possible Answers:

$-3cos^2(\pi+1)sin(\pi+1)$

$-3cos^2(\pi+1)$

$3cos^2(\pi+1)$

$3cos^2(\pi+1)sin(\pi+1)$

Correct answer:

$-3cos^2(\pi+1)sin(\pi+1)$

Explanation:

Let us first determine the derivative of f(x) by successive applications of the chain rule. The given function is a composition of three simpler functions: first add to , then take the cosine of the sum, and then cube the result. By the chain rule, we must differentiate with respect to each of these functions and take their product to obtain the derivative of the original function, as shown below:

f(x)=cos^3(x+1),

f'(x)=(3cos^2(x+1))(-sin(x+1))(1)

= -3cos^2(x+1)sin(x+1)

To determine the value of this function at $x=\pi$ , simply substitute $\pi$ for :

$-3cos^2(\pi+1)sin(\pi+1)$

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Example Question #12 : Other Derivative Review

Evaluate.

$\int \ \cos(2x ) \ dx$

Possible Answers:

$F(x) = \frac{\sin (2x)}{2}$

$F(x) = \frac{\cos (2x)}{2}$

$F(x) = - \frac{\sin (x)}{2}$

Answer not listed

$F(x) = - \frac{\cos (x)}{2}$

Correct answer:

$F(x) = \frac{\sin (2x)}{2}$

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = \cos(2x)$ .

The antiderivative is $F(x) = \frac{\sin (2x)}{2}$ .

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Example Question #13 : Other Derivative Review

Evaluate.

$\int \sqrt{x-2} \ dx$

Possible Answers:

$F(x) = - \frac{(x-2)^{\frac{3}{2}}}{3}$

$F(x) = (x-2)^{\frac{1}{2} }$

Answer not listed

$F(x) = (x-2)^{\frac{3}{2}}$

$F(x) = \frac{2(x-2)^{\frac{3}{2}}}{3}$

Correct answer:

$F(x) = \frac{2(x-2)^{\frac{3}{2}}}{3}$

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = \sqrt{x-2}$ and u = x-2 .

The antiderivative is $F(x) = \frac{2(x-2)^{\frac{3}{2}}}{3}$ .

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Example Question #14 : Other Derivative Review

Evaluate.

$\int e^{3x-1} \ dx$

Possible Answers:

Answer not listed

$F(x) = \frac{e^{3x-1}}{3}$

$F(x) = e^{3x-1}$

$F(x) = \frac{e^{x}}{3x-1}$

$F(x) =3 e^{3x-1}$

Correct answer:

$F(x) = \frac{e^{3x-1}}{3}$

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = e^{3x-1}$ and u = 3x-1 .

The antiderivative is $F(x) = \frac{e^{3x-1}}{3}$ .

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Example Question #15 : Other Derivative Review

Suppose and are related implicitly by the equation x^3+y+y^3=x . Find $\frac{d^2y}{dx^2}$ in terms of and .

Possible Answers:

$-\frac{3x^2-1}{3y^2+1}$

$\frac{3x^2-1}{3y^2+1}$

$\frac{6x(1+3y^2)^2+6y(1-3x^2)^2}{(1+3y^2)^3}$

$-\frac{6x(1+3y^2)^2+6y(1-3x^2)^2}{(1+3y^2)^3}$

$-\frac{6x(1+3y^2)^2-6y(1-3x^2)^2}{(1+3y^2)^3}$

Correct answer:

$-\frac{6x(1+3y^2)^2+6y(1-3x^2)^2}{(1+3y^2)^3}$

Explanation:

To take the second derivative of this implicit function, we must take the first derivative. To do so, we take the derivative of our implicit function with respect to :

$\frac{d}{dx}(x^3+y+y^3)=\frac{d}{dx}(x)$

$3x^2+\frac{dy}{dx}+3y^2\frac{dy}{dx}=1$

Here, we invoked the chain rule to take the derivatives of the terms, imagining as the inner function. This gives us an equation in $\frac{dy}{dx}$ which we can then solve for $\frac{dy}{dx}$ by algebraic means:

$\frac{dy}{dx}+3y^2\frac{dy}{dx}=1-3x^2$

$\frac{dy}{dx}(1+3y^2)=1-3x^2$

$\frac{dy}{dx}=\frac{1-3x^2}{1+3y^2}$

Now we have the value of $\frac{dy}{dx}$ . Now to find $\frac{d^2y}{dx^2}$ , we take the derivative of the above function with respect to :

$\frac{d^2x}{dy^2}=\frac{d}{dx}\biggr(\frac{1-3x^2}{1+3y^2}\biggr)$

$\frac{d^2x}{dy^2}=\frac{\big[\frac{d}{dx}(1-3x^2)\big](1+3y^2)-(1-3x^2)\big[\frac{d}{dx}(1+3y^2)\big]}{(1+3y^2)^2}$

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)-(1-3x^2)\big(6y\frac{dy}{dx}\big)}{(1+3y^2)^2}$

In the above, we used the quotient rule to take the derivative of the fraction, and then again using the chain rule on expressions involving . Now since our derivative must be in terms of and , we need to get rid of the $\frac{dy}{dx}$ in the above equation. Thus we substitute the first derivative expression $\frac{1-3x^2}{1+3y^2}$ which we found at the start of the problem into the fraction above:

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)-(1-3x^2)\bigg(6y\big(\frac{1-3x^2}{1+3y^2}\big)\bigg)}{(1+3y^2)^2}$

Now it is a matter of simplifying these equations. To do so, we multiply top and bottom by 1+3y^2 :

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)-(1-3x^2)\bigg(6y\big(\frac{1-3x^2}{1+3y^2}\big)\bigg)}{(1+3y^2)^2}\frac{1+3y^2}{1+3y^2}$

$\frac{d^2x}{dy^2}=\frac{-6x(1+3y^2)^2-(1-3x^2)6y(1-3x^2)}{(1+3y^2)^3}$

$\frac{d^2x}{dy^2}=-\frac{6x(1+3y^2)^2+6y(1-3x^2)^2}{(1+3y^2)^3}$

Finally, we arrive at the desired result.

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Example Question #1612 : Calculus Ii

Evaluate.

$\int x^4 \ dx$

Possible Answers:

$F(x) = \frac{1}{3}x^3$

Answer not listed

$F(x) = \frac{1}{4}x^4$

F(x) = x^4

$F(x) = \frac{1}{4}x$

Correct answer:

$F(x) = \frac{1}{4}x^4$

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, f(x) = x^4 .

The antiderivative is $F(x) = \frac{1}{4}x^4$ .

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Example Question #12 : Other Derivative Review

What is the derivative of tan(x) ?

Possible Answers:

sec(x)

$\frac{-1}{\sqrt{1-x^2}}+C$

$\frac{1}{\sqrt{1-x^2}}$

cot(x)

sec^2(x)

Correct answer:

sec^2(x)

Explanation:

The trigonometric functions have specific derivatives that one needs to memorize.

The basic trigonometric derivatives are as follows.

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

This particular question asks for the derivative for tangent and thus the correct solution is,

$tan(x)\frac{d}{dx}=sec^2(x)$

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Example Question #1621 : Calculus Ii

What is v(0) when x(t)=-4t^2+3t-1 ?

Possible Answers:

Correct answer:

Explanation:

To find the value of the velocity at 3, find the derivative of the position function. Remember, when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent.

Therefore,

$v(t)=x'(t)=2\cdot -4t^{2-1}+1\cdot 3t^{1-1}$

v(t)=-8t+3 .

Then, plug in 0 for t.

Therefore,

v(0)=-8(0)+3=3 .

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Example Question #1621 : Calculus Ii

Find $\frac{dy}{dx}$ by implicit differentiation

$y\sqrt{x}-x\sqrt{y}=16$

Possible Answers:

$\frac{2x\sqrt{y}-y}{2y\sqrt{x}-x}$

$\frac{2x\sqrt{x}-y}{2y\sqrt{y}-x}$

$\frac{2y\sqrt{x}-y}{2x\sqrt{y}-x}$

$\frac{2y\sqrt{y}-x}{2x\sqrt{x}-x}$

Correct answer:

$\frac{2y\sqrt{x}-y}{2x\sqrt{y}-x}$

Explanation:

Untitled

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Example Question #1621 : Calculus Ii

What is a possible function for f(x) if

$f^{\left( 4\right )}(x)=120(2x+3)$

Possible Answers:

f(x)=6x^5+120x^4

f(x)=2x^5+15x^4

f(x)=x^5+x^4

f(x)=2x^4+15x^5

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

Correct answer:

f(x)=2x^5+15x^4

Explanation:

Let f(x)=2x^5+15x^4

Step 1: Take the derivative of f(x) four times by using the power rule. The results are below:

f'(x)=10x^4+60x^3

f''(x)=40x^3+180x^2

$f^{(3)}=120x^2+360x$

$f^{(4)}=240x+360=120(2x+3)$

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Calculus 2 : Other Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Other Derivative Review

Example Question #12 : Other Derivative Review

Example Question #13 : Other Derivative Review

Example Question #14 : Other Derivative Review

Example Question #15 : Other Derivative Review

Example Question #1612 : Calculus Ii

Example Question #12 : Other Derivative Review

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

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