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Precalculus : Introductory Calculus

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Example Questions

Example Question #12 : Find The Second Derivative Of A Function

Find the second derivative of with respect to when

y=2x^6+3x^3-4x^2-5x+9 .

Possible Answers:

y''=60x^4+18x+8

y''=60x^4+18x-8

y''=13x^4+8x-6

y''=12x^5+9x^2-8x-5

Correct answer:

y''=60x^4+18x-8

Explanation:

For this problem we will need to use the power rule on each term.

The power rule is,

$y=x^n\rightarrow y'=nx^{n-1}$

Applying the power rule to our function we get the following derivative.

y=2x^6+3x^3-4x^2-5x+9

y'=12x^5+9x^2-8x-5+0

y''=60x^4+18x-8-0+0

y''=60x^4+18x-8

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Example Question #11 : Find The Second Derivative Of A Function

Find the second derivative of $y=\cos(x)$ .

Possible Answers:

$y''=-\sin(x)$

$y''=\cos(x)$

$y''=-\cos(x)$

$y''=\sin(x)$

Correct answer:

$y''=-\cos(x)$

Explanation:

We first need to find the first derivative of $y=\cos(x)$ . Remember that according to the derivatives of trigonometric functions, the derviative of cosine is negative sine and the derivative of sine is cosine.

Applying these rules we are able to find the first derivative.

$y'=-\sin(x)$

Now to find the second derivative we take the derivative of the first derivative.

$y''=-(\sin(x))'$

$y''=-(\cos(x))$

$y''=-\cos(x)$

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Example Question #703 : Pre Calculus

Find the second derivative of $f(x)=\ln(2+x^3)$ .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the second derivative of $f(x)=\ln(2+x^3)$ we will need to take the derivative of the first derivative.

To find the first derivative we will use the rule for natural logs which states,

$ln(u)'=\frac{1}{u}\cdot \frac{du}{dx}$

Applying this rule to our function we get the following.

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

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

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

Now that we have the first derivative we will take the derivative of it to get the second derivative. In order to do so we will need to use the quotient rule which states,

$\frac{f(x)}{g(x)}\rightarrow \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$

Applying this rule we get the following.

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

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

$f''(x)=\frac{12x+6x^4-9x^4}{(2+x^3)^2}$

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

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

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Example Question #12 : Find The Second Derivative Of A Function

Find the second derivative of $y=\sin(x)\cos(x)$ .

Possible Answers:

$y''=2\cos(x)-2\sin(x)$

$y''=2\cos(x)-\sin(x)-2\sin(x)\cos(x)$

$y''=(\cos(x))^2-(\sin(x))^2$

$y''=-4\sin(x)\cos(x)$

Correct answer:

$y''=-4\sin(x)\cos(x)$

Explanation:

To find the second derivative of this function we will need to use the following rules.

$\\ sin(x)\rightarrow cos(x)\\cos(x)\rightarrow-sin(x)$

Product Rule,

$f(x)g(x)\rightarrow f(x)g'(x)+g(x)f'(x)$ .

Chain Rule,

$[f(g(x))]'\rightarrow f'(g(x))\cdot g'(x)$ .

Applying these rules we can find both the first derivative and then the second derivative.

$y=\sin(x)\cos(x)$

$y'=\sin(x)'\cos(x)+\sin(x)\cos(x)'$

$y'=\cos(x)\cos(x)+\sin(x)(-\sin(x))$

$y'=(\cos(x))^2-(\sin(x))^2$

$y''=2\cos(x)(\cos(x))'-2\sin(x)(\sin(x))'$

$y''=2\cos(x)(-\sin(x))-2\sin(x)\cos(x)$

$y''=-2\sin(x)\cos(x)-2\sin(x)\cos(x)$

$y''=-4\sin(x)\cos(x)$

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Example Question #13 : Find The Second Derivative Of A Function

Find the second derivative of: 3x^4+2x^3+2x^2+4

Possible Answers:

f''(x)=12x^3+6x^2+4

f''(x)=36x^2+12x+4

f''(x)=12x^3+6x^2+4x

f''(x)=36x^2+12x+4x

None of the other answers.

Correct answer:

f''(x)=36x^2+12x+4

Explanation:

The derivative of this function, known as f'(x) (read as f prime of x), can be found by using the Power rule of derivatives on each term in the function. Power rule: $x^n=nx^{n-1}$ . If there is a value in front of x we multiply it by the "n" we carried over. For example taking the derivative of 2x^3=6x^2 . To find the second derivative, simply repeat the process. f''(x) is read as "f double prime of x" or "the second derivative of f(x)".

f(x)=3x^4+2x^3+2x^2+4

f'(x)=12x^3+6x^2+4x

f''(x)=36x^2+12x+4

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Example Question #21 : Find The Second Derivative Of A Function

Find the second derivative of the function

$h(x)=e^{x}sin(x)$

Possible Answers:

$h''(x)=2x^{2}e^{x}-sin(x)$

$h''(x)=2e^{}xcos(x)$

$h''(x)=e^{x}+cos(x)$

$h''(x)=e^{x}sin(x)+e^{x}cos(x)$

$h''(x)=2e^{x}sin(x)$

Correct answer:

$h''(x)=2e^{}xcos(x)$

Explanation:

Use the product rule to get the first derivative.

Let $f(x)=e^{x}$ and g(x)=sin(x)

h'(x)=f'(x)g(x)+g'(x)f(x)

$h'(x)=(e^{x})'sin(x)+e^{x}(sin(x))'$

$h'(x)=e^{x}sin(x)+e^{x}cos(x)$

Use the product rule again for the second derivative.

$h''(x)=e^{x}sin(x)+e^{x}cos(x)+e^{x}cos(x)-e^{x}sin(x)$

$h''(x)=2e^{}xcos(x)$

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Example Question #22 : Find The Second Derivative Of A Function

Find the second derivative of y=3x^4+5x^3 .

Possible Answers:

36x^3+30x^2

36x^4+30x^3

81x^3+15x^2

12x^3+15x^2

36x^2+30x

Correct answer:

36x^2+30x

Explanation:

To derive, use the power rule for derivatives.

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

Find the first derivative by taking the derivative of each term.

$y' = 4\cdot 3x^{4-1}+ 3\cdot 5x^{3-1} = 12x^3+15x^2$

Take the derivative of .

$y'' = 3\cdot 12x^{3-1}+2\cdot 15x^{2-1} = 36x^2+30x$

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Example Question #23 : Find The Second Derivative Of A Function

Determine the second derivative with the respect to x: $y=2e^{3x}$

Possible Answers:

$18x^2e^{3x-3}-6xe^{3x-3}$

$18 e^{9x}$

18e^x

$18 e^{3x-3}$

$18 e^{3x}$

Correct answer:

$18 e^{3x}$

Explanation:

To solve this, we first need to know the derivative of y=e^x with the respect to .

$\frac{d}{dx}e^x = e^x$

This problem will also involve the chain rule, which means that there we will need to take the derivative of the inner function inside e^x , since the power is not to the power of .

Find the first derivative.

$\frac{dy}{dx}= 2\cdot e^{3x} \cdot 3 = 6e^{3x}$

Find the second derivative by differentiating the first derivative.

$\frac{d^2y}{dx^2} = 6\cdot \e^{3x}\cdot 3 = 18 e^{3x}$

The answer is: $18 e^{3x}$

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Example Question #81 : Introductory Calculus

Find the second dervative for the following function.

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

Possible Answers:

x^2+2x+5

2x+2

Correct answer:

Explanation:

To find the second derivative, simply take the dervative twice according to the rules of derivatives.

f'(x)=2x+2

f''(x)=2

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Example Question #82 : Introductory Calculus

Find the second derivative of the following equation:

f(x)=0

Possible Answers:

undefined

Correct answer:

Explanation:

To solve, simply realize a constant derived is always 0. Thus, the answer is 0.

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Precalculus : Introductory Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #12 : Find The Second Derivative Of A Function

Example Question #11 : Find The Second Derivative Of A Function

Example Question #703 : Pre Calculus

Example Question #12 : Find The Second Derivative Of A Function

Example Question #13 : Find The Second Derivative Of A Function

Example Question #21 : Find The Second Derivative Of A Function

Example Question #22 : Find The Second Derivative Of A Function

Example Question #23 : Find The Second Derivative Of A Function

Example Question #81 : Introductory Calculus

Example Question #82 : Introductory Calculus

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