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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #141 : First And Second Derivatives Of Functions

What is the first derivative of the following equation?

$y=(x^3+2x)^{cosx}$

Possible Answers:

$\frac{dy}{dx}=(x^{3}+2x)^{sinx}(\frac{cosx(3x^{2}+2)}{x^{3}+2x}-sinxln(x^{3}+2x))$

$\frac{dy}{dx}=(x^{3}+2x)^{cosx}(\frac{cosx(3x^{2}+2)}{x^{3}+2x})$

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

$\frac{dy}{dx}=(x^{3}+2x)^{cosx}[(\frac{cosx(3x^{2}+2)}{x^{3}+2x}-sinxln(x^{3}+2x)]$

Correct answer:

$\frac{dy}{dx}=(x^{3}+2x)^{cosx}[(\frac{cosx(3x^{2}+2)}{x^{3}+2x}-sinxln(x^{3}+2x)]$

Explanation:

To solve this problem, we have to use several tricks. First, we take the natural log of each side, so we can bring down the exponent on the right side of the equation, which looks like this:

$lny=cosxln(x^{3}+2x)$ . Now, we differentiate each side of the equation, keeping in mind that we need to use implicit differentiation for the left side:

$\frac{dy}{dx}\frac{1}{y}=\frac{d}{dx}cosxln(x^{3}+2x)$ . To differentiate the right side, we use a combination of the chain and product rules, which looks like this:

$\frac{dy}{dx}\frac{1}{y}=\frac{d}{dx}cosxln(x^{3}+2x)+cosx\frac{d}{dx}ln(x^{3}+2x)$ , which becomes:

$\frac{dy}{dx}\frac{1}{y}=-sinxln(x^{3}+2x)+cosx(\frac{1}{x^{3}+2x})(3x^{2}+2)$ . This simplifies further to: $\frac{dy}{dx}\frac{1}{y}=(\frac{(cosx)(3x^{2}+2)}{x^{3}+2x})-sinxln(x^{3}+2x)$ . Lastly, we multiple both sides of the equation by . This gives us:

$\frac{dy}{dx}=y(\frac{(cosx)(3x^{2}+2)}{x^{3}+2x})-sinxln(x^{3}+2x)$ . Looking back at our original equation gives our value for . Plugging in this value, we get our answer of:

$\frac{dy}{dx}=(x^3+2x)^{cosx}[(\frac{(cosx)(3x^{2}+2)}{x^{3}+2x})-sinxln(x^{3}+2x)]$

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Example Question #141 : First And Second Derivatives Of Functions

What is the second derivative of f(x)=4x^4-3x^3+2x^2-x ?

Possible Answers:

$f{}''(x)=48x^2-18x-5$

$f{}''(x)=48x^3-18x+4$

$f{}''(x)=48x^2-18x+4$

$f{}''(x)=24x^2-18x+4$

$f{}''(x)=48x^2+18x+4$

Correct answer:

$f{}''(x)=48x^2-18x+4$

Explanation:

Before you can take the second derivative, find the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=16x^3-9x^2+4x-1$

Now, take the second derivative from the first derivative:

$f{}''(x)=48x^2-18x+4$

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Example Question #341 : Derivative Review

Find the derivative of f(x)=14x^2-3x+1 .

Possible Answers:

f{}'(x)=28x+3

f{}'(x)=28x-3

$f{}'(x)=28x^2-3$

f{}'(x)=24x-3

$f{}'(x)=28x-3x^2$

Correct answer:

f{}'(x)=28x-3

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

f{}'(x)=28x-3

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Example Question #1462 : Calculus Ii

What is the second derivative of f(x)=9x^3+4x^2-2x+9 ?

Possible Answers:

f{}''(x)=24x+8

f{}''(x)=54x-8

f{}''(x)=54x+18

f{}''(x)=54x

f{}''(x)=54x+8

Correct answer:

f{}''(x)=54x+8

Explanation:

To take the derivative, multiply the exponent by the coefficient in front of the x term and then also subtract one from the exponent. Take the first derivative:

$f{}'(x)=27x^2+8x-2$

Now, take the second derivative from the first derivative:

f{}''(x)=54x+8

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Example Question #344 : Derivative Review

If f(x)=3x^3 - 6x^2 + 2x - 1 , what is $\frac{d^2 u}{dx^2}$ ?

Possible Answers:

$\frac{x^4}{4}-2x^3+x^2-x$

18x-12

x^2-6x+2

9x^2-12x+2

Correct answer:

18x-12

Explanation:

To find the second derivative, first one has to find the first derivative, then take the derivative of this result.

The derivative of f(x)=3x^3 - 6x^2 + 2x - 1 is 9x^2-12x+2 .

The derivative of 9x^2-12x+2 is 18x-12 , and this is our final answer.

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Example Question #342 : Derivative Review

If f(x)=x^2-8x+15 , what is $\frac{d^2 u}{dx^2}$ ?

Possible Answers:

4x-8

$\frac{x^3}{3}-4x^2+15x$

2x-8

Correct answer:

Explanation:

To find the second derivative, first one has to find the first derivative, then take the derivative of this result.

The derivative of f(x)=x^2-8x+15 is 2x-8 .

The derivative of 2x-8 is , and this is our final answer.

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Example Question #1472 : Calculus Ii

What is the second derivative of f(x)=4x^3-6x^2+x-4 ?

Possible Answers:

f{}''(x)=24x-12

f{}''(x)=24x

f{}''(x)=24x-1

$f{}''(x)=12x^2-12x+1$

f{}''(x)=2x-12

Correct answer:

f{}''(x)=24x-12

Explanation:

First, you have to take the first derivative. Multiply the exponent by the coefficient in front of the x term and then also subtract one from the exponent:

$f{}'(x)=12x^2-12x+1$

Now, take the second derivative from the first derivative, using the same process:

f{}''(x)=24x-12

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Example Question #1473 : Calculus Ii

What is the derivative of $f(x)=4x^{\frac{3}{2}}+2x$ ?

Possible Answers:

$f{}'(x)=6\sqrt{x}$

$f{}'(x)=6\sqrt{x}-2$

$f{}'(x)=6\sqrt{x}+12$

$f{}'(x)=\sqrt{x}+2$

$f{}'(x)=6\sqrt{x}+2$

Correct answer:

$f{}'(x)=6\sqrt{x}+2$

Explanation:

Recall that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

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

Simplify to get your answer:

$f{}'(x)=6\sqrt{x}+2$

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Example Question #1474 : Calculus Ii

What is the first derivative of $f(x)=2x^7-\frac {1}{4x^3}$ ?

Possible Answers:

$\frac {56x^{10}+3}{4x^4}$

$14x^6-\frac {3}{4x^4}$

$14x^6+\frac {3}{4x^4}$

Correct answer:

$14x^6-\frac {3}{4x^4}$

Explanation:

Step 1: Take derivative of 2x^7
The derivative of 2x^7 is 14x^6 using the power rule which states $\frac{d}{dx}x^n=nx^{n-1}$ .

Step 2: We will use Quotient Rule on the fraction:

First, f(x) and g(x) :

f(x)=1, g(x)=4x^3

Second, find f'(x) and g'(x) :

f'(x)=0, g(x)=12x^2

Use the formula:

$\frac {f'g-g'f}{g^2}=\frac {(0)(4x^3)-(12x^2)(1)}{(4x^3)^2}=\frac {-12x^2}{16x^6}=\frac {-3}{4x^4}$

Step 3: Take the derivatives from step 1 and step 2 and add them up..

The derivative of f(x) is

$14x^6-\frac {3}{4x^4}$ .

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Example Question #1475 : Calculus Ii

What is the derivative of $f(x)=3x^{\frac{3}{2}}$ ?

Possible Answers:

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

$f{}'(x)=\frac{1}{2}\sqrt{x}$

$f{}'(x)=\frac{9}{2}x^{\frac{5}{2}}$

$f{}'(x)=\frac{7}{2}\sqrt{x}$

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

Correct answer:

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

Explanation:

To take the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

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

Therefore, your answer is:

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

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #141 : First And Second Derivatives Of Functions

Example Question #141 : First And Second Derivatives Of Functions

Example Question #341 : Derivative Review

Example Question #1462 : Calculus Ii

Example Question #344 : Derivative Review

Example Question #342 : Derivative Review

Example Question #1472 : Calculus Ii

Example Question #1473 : Calculus Ii

Example Question #1474 : Calculus Ii

Example Question #1475 : Calculus Ii

All Calculus 2 Resources

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