Finding Derivative at a Point - Math

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Question

Find if the function is given by

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Answer

To find the derivative at , we first take the derivative of . By the derivative rule for logarithms,

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

Plugging in , we get

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

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Question

Find the derivative of the following function at the point .

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Answer

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

Now we plug in

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Question

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

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Answer

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:

In order to find the derivative of , 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$)\cdot2x$

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

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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Question

Find if the function is given by

Tap to reveal answer

Answer

To find the derivative at , we first take the derivative of . By the derivative rule for logarithms,

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

Plugging in , we get

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

← Didn't Know|Knew It →

Question

Find the derivative of the following function at the point .

Tap to reveal answer

Answer

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

Now we plug in

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Question

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

Tap to reveal answer

Answer

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:

In order to find the derivative of , 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$)\cdot2x$

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

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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Question

Find if the function is given by

Tap to reveal answer

Answer

To find the derivative at , we first take the derivative of . By the derivative rule for logarithms,

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

Plugging in , we get

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

← Didn't Know|Knew It →

Question

Find the derivative of the following function at the point .

Tap to reveal answer

Answer

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

Now we plug in

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

In order to find the derivative of , 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$)\cdot2x$

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

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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Question

Find if the function is given by

Tap to reveal answer

Answer

To find the derivative at , we first take the derivative of . By the derivative rule for logarithms,

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

Plugging in , we get

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

← Didn't Know|Knew It →

Question

Find the derivative of the following function at the point .

Tap to reveal answer

Answer

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

Now we plug in

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

In order to find the derivative of , 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$)\cdot2x$

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

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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