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High School Math : High School Math

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Example Questions

Example Question #35 : Derivatives

What is the second derivative of x^3+2x^2+5 ?

Possible Answers:

6x+4

$\frac{2}{3}x+\frac{1}{4}$

3x^2+4x

Correct answer:

6x+4

Explanation:

To find the second derivative, we need to start by finding the first derivative.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as 5x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})+(0*5x^{0-1})$

Notice that $(0*5x^{0-1}) =0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{3-1})+(2*2x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3*x^{2})+(2*2x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=3x^2+4x$

Now we repeat the process but using 3x^2+4x as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(2*3x^{2-1})+(1*4x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(2*3x^{1})+(1*4x^{0})$

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=(6x)+(4(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+4x)=6x+4$

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Example Question #36 : Derivatives

What is the second derivative of $4x^2+\frac{2}{3}x$ ?

Possible Answers:

$8x+\frac{2}{3}$

Correct answer:

Explanation:

To find the second derivative, we need to start by finding the first derivative.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

Now we repeat the process, but we use $8x+\frac{2}{3}$ as our expression.

For this problem, we're going to say that $\frac{2}{3}=\frac{2}{3}x^0$ since, as stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=(1*8x^{1-1})+(0*\frac{2}{3}x^{0-1})$

Notice that $(0*\frac{2}{3}x^{0-1})=0$ as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=(1*8x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=(8x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+\frac{2}{3})=8$

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Example Question #1 : Finding Second Derivative Of A Function

What is the second derivative of (x+6)^2 ?

Possible Answers:

$\frac{1}{2}$

6x+12

2x+12

Correct answer:

Explanation:

To find the second derivative, we need to start with the first derivative.

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: $\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))*g'(x)$

Plug in our equations.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6)^{1})*(1+0)$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6))$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=2x+12$

From here, we can use our normal power rule to find the second derivative.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(1*2x^{1-1})+(0*12x^{0-1})$

Anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(1*2x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=(2x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+12)=2$

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Example Question #2121 : High School Math

What is the second derivative of 4x^2+5x-3 ?

Possible Answers:

y=32x+5x^2

y=8

y=3

y=16x

y=8x

Correct answer:

y=8

Explanation:

To find the second derivative, we need to find the first derivative first. To find the first derivative, we can use the power rule.

For each variable, multiply by the exponent and reduce the exponent by one:

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(2*4x^{2-1})+(1*5x^{1-1})-(0*3x^0-1)$

Treat as 3x^0 since anything to the zero power is one.

Remember, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=(8x^{1})+(5x^{0})-0$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+5x-3)=8x+5$

Now follow the same process but for 8x+5 .

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(1*8x^{1-1})+(0*5x^{0-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=(1*8x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x+5)=8$

Therefore the second derivative will be the line y=8 .

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Example Question #1 : Finding Second Derivative Of A Function

Let f(x)=sin(x)-cos(x) .

Find the second derivative of f(x) .

Possible Answers:

f''(x)=sin^2(x)-cos^2(x)

f''(x)=cos(x)-sin(x)

f''(x)=sin^2(x)+cos^2(x)

f''(x)=sin(x)+cos(x)

f''(x)=sin(x)-cos(x)

Correct answer:

f''(x)=cos(x)-sin(x)

Explanation:

The second derivative is just the derivative of the first derivative. So first we find the first derivative of f(x) . Remember the derivative of $\sin x$ is $\cos x$ , and the derivative for $\cos x$ is $-\sin x$ .

f'(x)= cos(x)+sin(x)

Then to get the second derivative, we just derive this function again. So

f''(x)=-sin(x)+cos(x)=cos(x)-sin(x)

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Example Question #2 : Finding Second Derivative Of A Function

Define $f (x) = 6x^{3} - 12x^{2} + 4x -8$ .

What is f ' ' (x) ?

Possible Answers:

f ' '(x) = 36x - 24

f ' '(x) = 6x - 1 2

f ' '(x) = 36x

f ' '(x) = 18

f ''(x) = 18x - 24

Correct answer:

f ' '(x) = 36x - 24

Explanation:

Take the derivative of , then take the derivative of .

$f (x) = 6x^{3} - 12x^{2} + 4x -8$

$f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0$

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

$f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0$

f ' '(x) = 36x - 24

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Example Question #301 : Computation Of The Derivative

Define $g(x) = 7x^{5} - 6x^{3} - 17x$ .

What is g ''(x) ?

Possible Answers:

$g'' (x) = 70x^{3} - 36x$

$g'' (x) = 140x^{3} - 36x -17$

$g'' (x) = 70x^{3} - 18x$

$g'' (x) = 140x^{3} - 36x$

$g'' (x) = 140x^{3} - 18x$

Correct answer:

$g'' (x) = 140x^{3} - 36x$

Explanation:

Take the derivative of , then take the derivative of .

$g(x) = 7x^{5} - 6x^{3} - 17x$

$g' (x) = 5\cdot 7x^{5-1} - 3\cdot 6x^{3-1} - 17$

$g' (x) = 35x^{4} - 18x^{2} - 17$

$g'' (x) = 4\cdot 35x^{4-1} - 2\cdot 18x^{2-1} - 0$

$g'' (x) = 140x^{3} - 36x$

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Example Question #54 : Calculus I — Derivatives

Define $g(x) = \cos \left (x^{2} \right )$ .

What is g''(x) ?

Possible Answers:

$g''(x) =-4x^2 \cos(x^2)$

$g''(x) = -4x^2 \sin(x^2)$

$g''(x) =4x^2 \cos(x^2)$

$g''(x) = - 2 \cos(x^2)-4x^2 \sin(x^2) \right ]$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Correct answer:

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Explanation:

Take the derivative of , then take the derivative of .

$g(x) = \cos \left (x^{2} \right )$

$g'(x) = 2x \cdot [-\sin (x^2)]$

$g'(x) = - 2x \sin (x^2)$

$g''(x) = - 2\left [ 1 \cdot \sin(x^2)+ x \cdot 2x \cos(x^2) \right ]$

$g''(x) = - 2\left [ \sin(x^2)+ 2x^2 \cos(x^2) \right ]$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

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Example Question #55 : Calculus I — Derivatives

Define $f(x) = \frac{1}{e^{4x}}$ .

What is f ''(x) ?

Possible Answers:

$f''(x) = \frac{16 }{e^{4x}}$

$f''(x) = \frac{1 }{e^{4x-2}}$

$f''(x) =- \frac{16 }{e^{4x}}$

$f''(x) = \frac{1 }{e^{4x+2}}$

$f(x) = \frac{1}{16e^{4x}}$

Correct answer:

$f''(x) = \frac{16 }{e^{4x}}$

Explanation:

Rewrite:

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

$f(x) = e^{-4x}$

Take the derivative of , then take the derivative of .

$f'(x) = -4 e^{-4x}$

$f''(x) = -4 \cdot \left (-4 \right )e^{-4x} = 16e^{-4x} = \frac{16 }{e^{4x}}$

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Example Question #56 : Calculus I — Derivatives

What is the second derivative of x^2+5x ?

Possible Answers:

2x+5

3x^2+5

$\frac{1}{3}x^3+\frac{5}{2}x^2+c$

Correct answer:

Explanation:

To get the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2*x^{2-1})+(1*5x^{1-0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5x^0$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5$

Now we do the same process again, but using 2x+5 as our expression:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(1*2x^{1-1})+(0*5x^{0-1})$

Notice that $(0*5x^{0-1})=0$ , as anything times zero will be zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(1*2x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(2x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(2*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=2$

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #35 : Derivatives

Example Question #36 : Derivatives

Example Question #1 : Finding Second Derivative Of A Function

Example Question #2121 : High School Math

Example Question #1 : Finding Second Derivative Of A Function

Example Question #2 : Finding Second Derivative Of A Function

Example Question #301 : Computation Of The Derivative

Example Question #54 : Calculus I — Derivatives

Example Question #55 : Calculus I — Derivatives

Example Question #56 : Calculus I — Derivatives

All High School Math Resources

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