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AP Calculus AB : Derivative defined as the limit of the difference quotient

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Example Questions

Example Question #4 : Finding Derivatives

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

Possible Answers:

x^2

$\frac{1}{2}x+5$

2x+5

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

Correct answer:

2x+5

Explanation:

To solve this problem, 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.

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

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

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Example Question #2 : Finding Derivative Of A Function

What is the derivative of 5x+8 ?

Possible Answers:

5x^2+8x+c

$\frac{5}{2}x^2+8x$

Correct answer:

Explanation:

To solve this problem, 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 is 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.

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

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

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Example Question #6 : General Derivatives And Rules

What is the derivative of ?

Possible Answers:

$\frac{3}{2}x$

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

2+c

Correct answer:

Explanation:

To get $\frac{\mathrm{d} }{\mathrm{d} x}(2x)$ , we can use the power rule.

Since the exponent of the is , as 2x=2x^1 , we lower the exponent by one and then multiply the coefficient by that original exponent:

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

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

Anything to the power is .

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

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

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Example Question #3 : Concept Of The Derivative

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

Possible Answers:

20x

12x

Correct answer:

12x

Explanation:

To solve this equation, 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.

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

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

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

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Example Question #4 : Finding Derivative Of A Function

What is the derivative of x+1 ?

Possible Answers:

$\frac{1}{x}$

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

Correct answer:

Explanation:

To solve this equation, 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:

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

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Example Question #1 : Derivative Defined As The Limit Of The Difference Quotient

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

Possible Answers:

24x+13

6x^2+13

2x+1

24x+1

Correct answer:

24x+13

Explanation:

To solve this equation, 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.

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

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

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Example Question #31 : Calculus I — Derivatives

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

Possible Answers:

$8x^2+\frac{1}{2}x$

$8x^{3}+\frac{1}{2}$

4x+1

$\frac{2}{5}x^5+\frac{1}{4}x^2+c$

Correct answer:

$8x^{3}+\frac{1}{2}$

Explanation:

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.

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

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

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Example Question #32 : Calculus I — Derivatives

What is the derivative of 3x+12 ?

Possible Answers:

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

3x^2

Correct answer:

Explanation:

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.

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

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

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Example Question #2 : Derivative Defined As The Limit Of The Difference Quotient

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

Possible Answers:

2x+12

10x+12

2x+1

Correct answer:

10x+12

Explanation:

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.

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

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

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Example Question #3 : Derivative Defined As The Limit Of The Difference Quotient

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

Possible Answers:

16x

-16x

Correct answer:

-16x

Explanation:

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.

For this problem that would look like this:

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

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

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

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

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AP Calculus AB : Derivative defined as the limit of the difference quotient

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #4 : Finding Derivatives

Example Question #2 : Finding Derivative Of A Function

Example Question #6 : General Derivatives And Rules

Example Question #3 : Concept Of The Derivative

Example Question #4 : Finding Derivative Of A Function

Example Question #1 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #31 : Calculus I — Derivatives

Example Question #32 : Calculus I — Derivatives

Example Question #2 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #3 : Derivative Defined As The Limit Of The Difference Quotient

All AP Calculus AB Resources

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