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High School Math : Calculus I — Derivatives

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Example Questions

Example Question #15 : Finding Derivatives

$\frac{\mathrm{d} }{\mathrm{d} x}(4x+3)=?$

Possible Answers:

2x^2+3x

4x^2+3x+c

Correct answer:

Explanation:

To find the derivative of the problem, we can use the power rule. The power rule says to multiply the coefficient of the variable by the exponent of the variable and then lower the exponent value by one.

To make that work, we're going to treat as 3x^0 , since anything to the zero power is one.

This means that 4x+3 is the same as 4x^1+3x^0 .

Now use the power rule:

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

Anything times zero is zero.

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

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

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

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Example Question #16 : Finding Derivatives

What is the first derivative of x^4-3x^2+8 ?

Possible Answers:

$4x^{3}-6x+\frac{8}{x}$

$4x^{3}-6x$

12x^2+6

Correct answer:

$4x^{3}-6x$

Explanation:

To find the derivative of x^4-3x^2+8 , we can use the power rule.

The power rule states that we multiply each variable by its current exponent and then lower the exponent of each variable by one.

Since x^0=1 , we're going to treat as 8x^0 .

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

Anything times zero is zero, so our final term $(0*8x^{0-1})=0$ , regardless of the power of the exponent.

Simplify what we have.

$\frac{\mathrm{d} }{\mathrm{d} x}x^4-3x^2+8x^0=(4x^{3})-(6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}x^4-3x^2+8x^0=4x^{3}-6x$

Our final solution, then, is 4x^3-6x .

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Example Question #21 : Finding Derivatives

If $f(x)=\frac{2}{3}x^3+7x^2-12x$ , what is f'(x) ?

Possible Answers:

$\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

There is no f'(x) for this equation.

$2x^{2}+14x-12$

4x+14

Correct answer:

$2x^{2}+14x-12$

Explanation:

For this problem, we can use the power rule. The power rule states that we multiply each variable by its current exponent and then lower that exponent by one.

$\frac{\mathrm{d} }{\mathrm{d} x}f(x)=(3*\frac{2}{3}x^{3-1})+(2*7x^{2-1})-(1*12x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}f(x)=(2x^{2})+(14x^{1})-(12x^{0})$

Anything to the zero power is one, so 12x^0=12(1)=12 .

Therefore, $f'(x)=2x^{2}+14x-12$ .

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Example Question #41 : Calculus I — Derivatives

Find the derivative of the following function:

f(x) = 6x^2+4x + 7

Possible Answers:

f'(x)= 6x + 4

f'(x)= 12x + 7

f'(x)= 12x^2 + 7

f'(x)= 12x+4

f'(x) = 6x

Correct answer:

f'(x)= 12x+4

Explanation:

We use the power rule on each term of the function.

The first term

6x^2

becomes

$2 \times 6 x^2 = 12 x$ .

The second term

becomes

The final term, 7, is a constant, so its derivative is simply zero.

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Example Question #1 : Finding Second Derivative Of A Function

What is the second derivative of $2x^4+\frac{1}{2}x$ ?

Possible Answers:

$24x^{2}+\frac{1}{2x}$

$24x^{-2}$

$24x^{2}$

Correct answer:

$24x^{2}$

Explanation:

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

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

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

Simplify.

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

Remember that anything to the zero power is equal to one.

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

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

Now we repeat the process, but using $8x^3+\frac{1}{2}$ as our expression.

We're going to treat $\frac{1}{2}$ as being $\frac{1}{2}x^0$ since anything to the zero power is equal to one.

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

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=24x^{2}$

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Example Question #42 : Calculus I — Derivatives

What is the second derivative of 3x+12 ?

Possible Answers:

Undefined

x+8

$\frac{3}{2}x^2$

Correct answer:

Explanation:

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

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

We are going to treat as 12x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(0*12x^{0-1})$

Notice that $(0*12x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3$

Now we repeat the process but use , or 3x^0 , as our expression.

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

As stated before, anything times zero is zero.

Therefore, $\frac{\mathrm{d} }{\mathrm{d} x}(3x^0)=0$ .

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Example Question #3 : Finding Second Derivative Of A Function

What is the second derivative of 5x^2+12x ?

Possible Answers:

10x+12

Correct answer:

Explanation:

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

Now we repeat the process using 10x+12 as our expression.

We're going to treat as 12x^0 .

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})+(0*12x^{0-1})$

Notice that $(0*12x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(10x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10$

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Example Question #4 : Finding Second Derivative Of A Function

What is the second derivative of -8x^2+15 ?

Possible Answers:

16x

Correct answer:

Explanation:

To find the second derivative, we start by taking the first derivative.

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

We're going to treat as 15x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $(0*15x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

Now we repeat the process, but using -16x as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$

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Example Question #5 : Finding Second Derivative Of A Function

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

Possible Answers:

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

3x^2+4x

6x+4

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 #43 : Calculus I — 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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High School Math : Calculus I — Derivatives

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #15 : Finding Derivatives

Example Question #16 : Finding Derivatives

Example Question #21 : Finding Derivatives

Example Question #41 : Calculus I — Derivatives

Example Question #1 : Finding Second Derivative Of A Function

Example Question #42 : Calculus I — Derivatives

Example Question #3 : Finding Second Derivative Of A Function

Example Question #4 : Finding Second Derivative Of A Function

Example Question #5 : Finding Second Derivative Of A Function

Example Question #43 : Calculus I — Derivatives

All High School Math Resources

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