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AP Calculus AB : Computation of the Derivative

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Example Question #54 : 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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Example Question #55 : Calculus I — Derivatives

What is the second derivative of 5x+8 ?

Possible Answers:

$\frac{5}{x}$

$\frac{1}{5x}$

Undefined

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.

We're going to treat as 8x^0 , as anything to the zero power is one.

That means this problem will look like this:

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

Notice that $(0*8x^{0-1})=0$ as anything times zero will be zero.

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

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

Remember, anything to the zero power is one.

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

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

Now to get the second derivative we repeat those steps, but instead of using 5x+8 , we use 5x^0 .

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

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

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

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Example Question #31 : General Derivatives And Rules

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

Possible Answers:

12x+5

3x^2+2

18x^4+13x^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.

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

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

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

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

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

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

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

Just like before, we're going to treat as 2x^0 .

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

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

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

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

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Example Question #31 : General Derivatives And Rules

If f(x)=6x^2+8 , what is f''(x) ?

Possible Answers:

12x

Correct answer:

Explanation:

The question is asking us for the second derivative of the equation. First, we need to find the first derivative.

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as 8x^0 since anything to the zero power is one.

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

Notice that $0*8x^{0-1}=0$ since anything times zero is zero.

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

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

Now we do the exact same process but using 12x as our expression.

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

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

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

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

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

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Example Question #62 : Calculus I — Derivatives

What is the second derivative of x+1 ?

Possible Answers:

Undefined

$-\sqrt x$

Correct answer:

Explanation:

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

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as 1x^0 since anything to the zero power is one.

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

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

That leaves us with $\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})$ .

Simplify.

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

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

As stated earlier, anything to the zero power is one, leaving us with:

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

Now we can repeat the process using or x^0 as our equation.

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

As pointed out before, anything times zero is zero, meaning that $\frac{\mathrm{d} }{\mathrm{d} x}(x^0)=0$ .

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Example Question #63 : Calculus I — Derivatives

What is the second derivative of 12x^2+13x+4 ?

Possible Answers:

144

24x+13

24 x

Correct answer:

Explanation:

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

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as 4x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})+(0*4x^{0-1})$

Notice that $(0*4x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{1})+(1*13x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

Now we repeat the process using 24x+13 as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(1*24x^{1-1})+(0*13x^{0-1})$

Like before, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+10)=(1*24x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=24$

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Example Question #51 : Calculus I — Derivatives

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

What is g ''(x) ?

Possible Answers:

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

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

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

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

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

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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AP Calculus AB : Computation of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #54 : Calculus I — Derivatives

Example Question #55 : Calculus I — Derivatives

Example Question #31 : General Derivatives And Rules

Example Question #31 : General Derivatives And Rules

Example Question #62 : Calculus I — Derivatives

Example Question #63 : Calculus I — Derivatives

Example Question #51 : Calculus I — Derivatives

All AP Calculus AB Resources

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