High School Math : Finding Derivative at a Point

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : General Derivatives And Rules

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

\(\displaystyle f(x) = ln(x)\)

Possible Answers:

\(\displaystyle \frac{3}{2}\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle 0\)

\(\displaystyle -1\)

\(\displaystyle 1\)

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 #2 : 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:

\(\displaystyle 109\)

\(\displaystyle 110\)

\(\displaystyle 106\)

\(\displaystyle 108\)

\(\displaystyle 107\)

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 #3 : 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:

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

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

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

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

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

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

 

 

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