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High School Math : Finding Derivative at a Point

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Example Questions

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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High School Math : Finding Derivative at a Point

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

All High School Math Resources

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