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High School Math : General Derivatives and Rules

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Example Question #21 : Calculus I — Derivatives

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

f(x) = ln(x)

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

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

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

Plugging in x=2 , we get

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

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Example Question #22 : Calculus I — Derivatives

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

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

Possible Answers:

106

108

109

110

107

Correct answer:

109

Explanation:

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

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

Now we plug in x=3

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

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Example Question #23 : Calculus I — Derivatives

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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

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

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

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

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

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

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

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

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

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Example Question #1 : Finding Derivative Of A Function

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

Possible Answers:

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

$8\tfrac{2}{3}x^2$

Correct answer:

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

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.

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

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

Remember that anything to the zero power is one.

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

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

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Example Question #2 : Finding Derivative Of A Function

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

Possible Answers:

8x-8

16x+5

8x+5

Correct answer:

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:

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

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

Remember, anything times zero is zero.

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

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

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Example Question #1 : Finding Derivatives

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

Possible Answers:

192

256

Correct answer:

192

Explanation:

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

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

Substitute:

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

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

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

Possible Answers:

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

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

2x+5

x^2

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 #5 : Finding Derivative Of A Function

What is the derivative of 5x+8 ?

Possible Answers:

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

5x^2+8x+c

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 : Finding Derivative Of A Function

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

Possible Answers:

3x^2+2x

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

3x^2+2

Correct answer:

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 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.

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

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

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

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Example Question #7 : Finding Derivative Of A Function

What is the derivative of ?

Possible Answers:

2+c

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

$\frac{3}{2}x$

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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High School Math : General Derivatives and Rules

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #21 : Calculus I — Derivatives

Example Question #22 : Calculus I — Derivatives

Example Question #23 : Calculus I — Derivatives

Example Question #1 : Finding Derivative Of A Function

Example Question #2 : Finding Derivative Of A Function

Example Question #1 : Finding Derivatives

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

Example Question #5 : Finding Derivative Of A Function

Example Question #6 : Finding Derivative Of A Function

Example Question #7 : Finding Derivative Of A Function

All High School Math Resources

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