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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #181 : Derivatives

Find the derivative of the following function at the point x=0 :

$f(x)=\frac{2x^3+x^2+e^x}{3x+1}$

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(x)=\frac{(3x+1)(6x^2+2x+e^x)-3(2x^3+x^2+e^x)}{(3x+1)^2}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}( \frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ ,
$\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ .

Now, plug in the point x=0 into the above function:

$f'(x)=\frac{(3(0)+1)(6(0)^2+2(0)+e^{(0)})-3(2(0)^3+(0)^2+e^{(0)})}{(3(0)+1)^2}$

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

$f'(0)=\frac{-2}{1}=-2$

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Example Question #1311 : Calculus Ii

Find the derivative of the following function at the point x=0 :

$f(x)=\tan(x)-\frac{3x}{4}$

Possible Answers:

$\frac{3}{4}$

$\frac{1}{4}$

$\frac{5}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

The derivative of the function is

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

and was found using the following rules:

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

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{^{n-1}}$

Now, plug in the given point into the first derivative function:

$f'(0)=1-\frac{3}{4}=\frac{1}{4}$

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Example Question #182 : Derivative Review

Find the derivative of the following function at x=0 :

$f(x)=x^3\cos(x)+\sin(3x)$

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

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

and was found using the following rules:

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

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, plug in the point we want into the derivative and solve:

f'(0)=0-0+3(1)=3

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Example Question #187 : Derivative Review

Evaluate the derivative at the point $\theta=\pi$ :

$f(\theta)=\cos^2(\theta)+e^\theta$

Possible Answers:

$2\pi$

$e^\pi$

Correct answer:

$e^\pi$

Explanation:

The derivative of the function is

$f'(\theta)=-2\cos(\theta)\sin(\theta)+e^\theta$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ .

Now, plug in the point asked into the above function and solve:

$f'(\pi)=0+e^\pi=e^\pi$

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Example Question #1312 : Calculus Ii

Find the derivative of the following function at $\theta=\pi$ :

$f(\theta)=\tan(\theta)+\theta e^\theta$

Possible Answers:

$e^\pi (\pi+1)-1$

$e^\pi (\pi+1)+1$

$e^\pi (\pi+1)$

$e^\pi (\pi-1)$

Correct answer:

$e^\pi (\pi+1)-1$

Explanation:

The derivative of the function is

$f'(\theta)=\sec^2(\theta)+e^\theta+\theta e^\theta$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

Finally, plug in $\pi$ for $\theta$ and we get our final answer:

$f'(\theta)=e^\pi (\pi+1)-1$

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Example Question #10 : Computation Of Derivatives

What is the rate of change of the function f(x)=1+2x+3x^4 at the point (1,6) ?

Possible Answers:

2594

Correct answer:

Explanation:

The rate of change of a function at a point is the value of the derivative at that point. First, take the derivative of f(x) using the power rule for each term.

Remember that the power rule is

$\frac{d}{dx}x^n=nx^{n-1}$ , and that the derivative of a constant is zero.

f'(x)=2+12x^3

Next, notice that the x-value of the point (1,6) is 1, so substitute 1 for x in the derivative.

f'(1)=2+12(1)^3=14

Therefore, the rate of change of f(x) at the point (1,6) is 14.

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Example Question #101 : Derivative At A Point

Calculate the derivative of f(x)=4x^2-x+2 at the point x=1 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x) .

Then, replace the value of x with the given point.

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

f(x)=4x^2-x+2

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

f'(x)=8x-1

Then, plug in the value of x and evaluate

f'(x=1)=8*1-1 = 7

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Example Question #1 : Derivative At A Point

Evaluate the first derivative if

f(x)=2e^x and x=0 .

Possible Answers:

f'(0)=0

f'(0)=2

f'(0)=1

f'(0)=e

Correct answer:

f'(0)=2

Explanation:

First we must find the first derivative of the function.

Because the derivative of the exponential function is the exponential function itelf, or

$\frac{\mathrm{d} }{\mathrm{d} x}[e^x]=e^x$

and taking the derivative is a linear operation,

we have that

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[f(x)]$

$=\frac{\mathrm{d} }{\mathrm{d} x}[2e^x]$

$=2\frac{\mathrm{d} }{\mathrm{d} x}[e^x]$

=2e^x

Now setting x=0

f'(0)=2e^0=2*1=2

Thus

f'(0)=2

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Example Question #1311 : Calculus Ii

What is the derivative of $(2+3cos(3x))^\pi$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Need to use the power rule which states: $\frac{d}{dx}u^n=nu^{n-1}\frac{du}{dx}$

In our problem $\frac{du}{dx}=-3sin(3x)$

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Example Question #1 : Computation Of Derivatives

$f (x) = 7x^{2} + 6x - 9$

Give f'(x) .

Possible Answers:

$f'(x) = 14x^{2}+6x$

f'(x) = 14x+6

f'(x) = 7x+6

f'(x) = 14x+12

f'(x) = 14x

Correct answer:

f'(x) = 14x+6

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

f ' (x) = 14x + 6

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #181 : Derivatives

Example Question #1311 : Calculus Ii

Example Question #182 : Derivative Review

Example Question #187 : Derivative Review

Example Question #1312 : Calculus Ii

Example Question #10 : Computation Of Derivatives

Example Question #101 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #1311 : Calculus Ii

Example Question #1 : Computation Of Derivatives

All Calculus 2 Resources

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