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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

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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 #16 : Other Derivative Review

Evaluate.

$\int x^4 \ dx$

Possible Answers:

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

F(x) = x^4

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

Answer not listed

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

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 #17 : Other Derivative Review

What is the derivative of tan(x) ?

Possible Answers:

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

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

sec^2(x)

cot(x)

sec(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 #11 : Other Derivative Review

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Untitled

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Example Question #1623 : 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

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

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

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

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

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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Example Question #1624 : Calculus Ii

$f(x)=\frac{5^{x^{2}}}{2^{x}}$ . Find f'(x) .

Possible Answers:

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{2^{x}}$

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(\frac{5}{2}))}{2^{x}}$

$f'(x)=\frac{x\cdot 5^{x^{2}}(\frac{1}{5}x-\frac{1}{2})}{2^{x}}$

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)+\ln(2))}{2^{x}}$

Correct answer:

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{2^{x}}$

Explanation:

Since both the numerator and denominator contain a variable, we must use the quotient rule.

$f'(x)=\frac{(2^{x}){\color{Green} \frac{\mathrm{d} }{\mathrm{d} x}(5^{x^{2}})}-(5^{x^{2}}){\color{Blue} \frac{\mathrm{d} }{\mathrm{d} x}(2^{x})}}{(2^{x})^{2}}$

Remember that the derivative of a constant raised to a variable power follows the pattern, $\frac{\mathrm{d} }{\mathrm{d} u}(a^{u})=a^{u}\ln\left | a \right | \mathrm{d} u$ , where a is a constant and u is a function of x.

Applying this we get,

$f'(x)= \frac{(2^{x}){\color{Green} (5^{x^{2}}(\ln5)(2x))}-(5^{x^{2}}){\color{Blue} (2^{x}(\ln2)(1))}}{(2^{x})^{2}}$

Now we simplify by factoring the greatest common factor out of the numerator.

$f'(x)=\frac{2^{x}\cdot 5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{(2^{x})^{2}}$

Then we can cancel a single $2^{x}$ from the numerator and denominator.

$f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{2^{x}}$

This is the correct answer.

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Example Question #1625 : Calculus Ii

Differentiate $f(t)=\frac{t}{1-e^t}$ .

Possible Answers:

$\frac{-te^t-1+e^t}{1-2e^t+e^{2t}}$

$\frac{1-e^t+te^t}{1-2e^t+e^{2t}}$

Undefined

e^t

$-\frac{1}{e^t}$

Correct answer:

$\frac{1-e^t+te^t}{1-2e^t+e^{2t}}$

Explanation:

Using the Quotient rule for derivatives, we know that the derivative will equal $\frac{(1-e^t)(1)-t(-e^t)}{(1-e^t)^2}$ .

When you simplify this, you get $\frac{1-e^t+te^t}{1-2e^t+e^{2t}}$ .

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #13 : Other Derivative Review

Example Question #14 : Other Derivative Review

Example Question #15 : Other Derivative Review

Example Question #16 : Other Derivative Review

Example Question #17 : Other Derivative Review

Example Question #1621 : Calculus Ii

Example Question #11 : Other Derivative Review

Example Question #1623 : Calculus Ii

Example Question #1624 : Calculus Ii

Example Question #1625 : Calculus Ii

All Calculus 2 Resources

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