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High School Math : Derivatives

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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 #64 : Calculus I — Derivatives

What is the second derivative of 4x+3 ?

Possible Answers:

2x^2+3x+c

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

Correct answer:

Explanation:

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

To find the first 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 value of the exponent 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.

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

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

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

Now we repeat the process, but using .

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

Since anything times zero is zero, $\frac{\mathrm{d} }{\mathrm{d} x}(4x^0)=0.$

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Example Question #65 : Calculus I — Derivatives

What is the second derivative of $\sin(2x)$ ?

Possible Answers:

$-\sin(2x)$

$2\cos(2x)$

$-\sin(4x^2)$

$2\sin(2x)$

$-4\sin(2x)$

Correct answer:

$-4\sin(2x)$

Explanation:

To start we need to find the first derivative. For that, we need to use the chain rule: f(g(x))=f'(g(x)*g'(x)

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

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

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

Repeat the process!

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

Remember that $\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$ :

$\frac{\mathrm{d} }{\mathrm{d} x}2\cos(2x)=-2\sin(2x)*2$

$\frac{\mathrm{d} }{\mathrm{d} x}2\cos(2x)=-4\sin(2x)$

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Example Question #51 : Derivatives

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

Possible Answers:

12x^2-6

18x^2

4x^3-6x

12x

Correct answer:

12x^2-6

Explanation:

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

To find the first 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 first derivative, then, is 4x^3-6x .

To find the second derivative, we repeat the process using 4x^3-6x for our equation.

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

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}4x^3-6x=(12x^{2})-(6x^{0})$

Remember that x^0=1 , which means our second derivative will be 12x^2-6 .

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Example Question #67 : Calculus I — Derivatives

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

Possible Answers:

$4x+4+\frac{12}{x}$

4x+14

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

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

Correct answer:

4x+14

Explanation:

To find f''(x) , or the second derivative of our function, we need to start by finding the first derivative.

To find the first derivative, 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, $f'(x)=2x^{2}+14x-12$ .

Now we repeat the process, but we use $f'(x)=2x^{2}+14x-12$ .

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

Remember, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}f'(x)=(4x^{1})+(14x^{0})-(0)$

f''(x)=4x+14

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Example Question #61 : Calculus I — Derivatives

If y= ln(x^3) what is the slope of the line at x=12 .

Possible Answers:

0.1

0.5

0.25

$\infty$

Correct answer:

0.25

Explanation:

The slope at any point on a line is also equal to the derivative. So first we want to find the derivative function of this function and then evaluate it at x=12 . So, to find the derivative we will need to use the chain rule. The chain rule says

$\frac{d}{dx}} f(g(x)) = f'(g(x))g'(x)$

so if we let f(x)= ln(x) and g(x)=x^3 then

since $f'(x)=\frac{1}{x}}$ and g'(x)=3x^2

$\frac{d}{dx}} f(g(x)) = f'(g(x))g'(x) = \frac{1}{x^3}*3x^2 = \frac{3}{x}$

Therefore we evaluate at x=12 and we get $\frac{3}{12}$ or .

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Example Question #1 : Using The Chain Rule

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

Possible Answers:

x+6

2x+13

4x+24

2x+12

Correct answer:

2x+12

Explanation:

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.

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

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

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

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Example Question #1 : Using The Chain Rule

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

Possible Answers:

$2\cos(2x)$

$-\frac{1}{2}\cos(2x)$

$2\cos(x)$

$\cos(2x)$

$2\sin(2x)$

Correct answer:

$2\cos(2x)$

Explanation:

For this problem we need to use the chain rule: f(g(x))=f'(g(x)*g'(x)

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

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

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

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High School Math : Derivatives

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #31 : General Derivatives And Rules

Example Question #62 : Calculus I — Derivatives

Example Question #63 : Calculus I — Derivatives

Example Question #64 : Calculus I — Derivatives

Example Question #65 : Calculus I — Derivatives

Example Question #51 : Derivatives

Example Question #67 : Calculus I — Derivatives

Example Question #61 : Calculus I — Derivatives

Example Question #1 : Using The Chain Rule

Example Question #1 : Using The Chain Rule

All High School Math Resources

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