High School Math : General Derivatives and Rules

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Finding Derivatives

Find \displaystyle f^{'}(2) if the function \displaystyle f(x) is given by

\displaystyle f(x) = ln(x)

Possible Answers:

\displaystyle \frac{1}{2}

\displaystyle \frac{3}{2}

\displaystyle -1

\displaystyle 1

\displaystyle 0

Correct answer:

\displaystyle \frac{1}{2}

Explanation:

To find the derivative at \displaystyle x=2, we first take the derivative of \displaystyle f(x). By the derivative rule for logarithms,

\displaystyle f'(x) = \frac{1}{x}

Plugging in \displaystyle x=2, we get

\displaystyle f'(2) = \frac{1}{2}

Example Question #1 : General Derivatives And Rules

Find the derivative of the following function at the point \displaystyle x=3.

\displaystyle f(x) = 4x^3+x+3

Possible Answers:

\displaystyle 109

\displaystyle 110

\displaystyle 108

\displaystyle 106

\displaystyle 107

Correct answer:

\displaystyle 109

Explanation:

Use the power rule on each term of the polynomial to get the derivative,

\displaystyle f'(x) = (3)(4)x^2+1

Now we plug in \displaystyle x=3

\displaystyle f'(3) = (12)(3)^2 + 1 = 109

Example Question #1 : General Derivatives And Rules

Let \displaystyle f(x)=x^2\sin(x^2). What is \displaystyle f'\left (\frac{\sqrt{\pi}}{2} \right )?

Possible Answers:

\displaystyle \frac{\sqrt{2\pi}(\pi + \sqrt{2})}{4}

\displaystyle \frac{\sqrt{2\pi}(\pi + 2)}{4}

\displaystyle \frac{(\pi + \sqrt{2})}{8}

\displaystyle \frac{\sqrt{2\pi}(\pi + 4)}{8}

\displaystyle \frac{{2\pi}(\sqrt{\pi} + 2)}{4}

Correct answer:

\displaystyle \frac{\sqrt{2\pi}(\pi + 4)}{8}

Explanation:

We need to find the first derivative of f(x). This will require us to apply both the Product and Chain Rules. When we apply the Product Rule, we obtain:

\displaystyle f'(x)=\sin(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]+x^2\cdot\frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]

In order to find the derivative of \displaystyle \sin(x^2), we will need to employ the Chain Rule.

 \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]=\cos(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]=\cos(x^2)\cdot2x

\displaystyle f'(x)=\sin(x^2)\cdot2x + x^2\cdot\cos(x^2)\cdot2x

 We can factor out a 2x to make this a little nicer to look at.

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

Now we must evaluate the derivative when x = \displaystyle \frac{\sqrt{\pi}}{2}.

\displaystyle f'\left (\frac{\sqrt{\pi}}{2} \right )=2\cdot\frac{\sqrt{\pi}}{2}(\sin\frac{\pi}{4}+\frac{\pi}{4}\cos\frac{\pi}{4})

\displaystyle =\sqrt{\pi}(\frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8})=\sqrt{2\pi}(\frac{1}{2}+\frac{\pi}{8})=\frac{\sqrt{2\pi}(\pi + 4)}{8}

The answer is \displaystyle \frac{\sqrt{2\pi}(\pi + 4)}{8}.

 

 

Example Question #23 : Calculus I — Derivatives

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

Possible Answers:

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

\displaystyle 3

\displaystyle 8

\displaystyle 3x

Correct answer:

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

Explanation:

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.

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

\displaystyle \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.

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

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

Example Question #3 : Finding Derivatives

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

Possible Answers:

\displaystyle 16x+5

\displaystyle 3

\displaystyle 8x+5

\displaystyle 2x

\displaystyle 8x-8

Correct answer:

\displaystyle 8x+5

Explanation:

This problem is best solved by using the power rule. For each variable, multiply by the exponent and reduce the exponent by one:

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

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

Remember, anything times zero is zero.

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

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

Example Question #3 : General Derivatives And Rules

Give the average rate of change of the function \displaystyle f(x) = 4 ^{x} on the interval  \displaystyle [3,4].

Possible Answers:

\displaystyle 256

\displaystyle 4

\displaystyle 192

\displaystyle 64

\displaystyle 16

Correct answer:

\displaystyle 192

Explanation:

The average rate of change of \displaystyle f on interval \displaystyle [a,b] is 

\displaystyle \frac{f (b ) - f(a)}{b-a}

Substitute:

\displaystyle \frac{f (4 ) - f(3)}{4-3} = \frac{4^{4} - 4^{3}}{1} = \frac{256 - 64}{1} = 192

Example Question #1 : Finding Derivative Of A Function

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

Possible Answers:

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

\displaystyle x^2

\displaystyle 2x+5

\displaystyle 7

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

Correct answer:

\displaystyle 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.

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

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

Remember that anything to the zero power is one.

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

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

Example Question #2 : General Derivatives And Rules

What is the derivative of \displaystyle 5x+8?

Possible Answers:

\displaystyle 13

\displaystyle 0

\displaystyle 5x^2+8x+c

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

\displaystyle 5

Correct answer:

\displaystyle 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.

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

That means this problem will look like this:

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

Notice that \displaystyle (0*8x^{0-1})=0, as anything times zero is zero.

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

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

Remember, anything to the zero power is one.

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

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

Example Question #2 : Finding Derivative Of A Function

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

Possible Answers:

\displaystyle 5

\displaystyle \frac{1}{3}x^2+\frac{1}{2}x+\frac{1}{5}

\displaystyle 3x^2+2

\displaystyle 3x^2+2x

\displaystyle 25

Correct answer:

\displaystyle 3x^2+2

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 \displaystyle 5 as \displaystyle 5x^0, as anything to the zero power is one.

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

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

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

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

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

Example Question #3 : Finding Derivatives

What is the derivative of \displaystyle 2x?

Possible Answers:

\displaystyle 2

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

\displaystyle x

\displaystyle 2+c

\displaystyle \frac{3}{2}x

Correct answer:

\displaystyle 2

Explanation:

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

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

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

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

Anything to the \displaystyle 0 power is \displaystyle 1.

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

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

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