Finding the Derivative - Complex Fraction with Radicals

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April 19th 2010, 05:15 PM
soad

Finding the Derivative - Complex Fraction with Radicals

So I've been doing good with these until this problem.

Find $f'(a)$

$f(x)=\frac{8}{\sqrt{x+2}}$

This is what we have to start with.

$\frac{\frac{8}{\sqrt{a+h+2}}-\frac{8}{\sqrt{a+2}}}{h}$

I began multiplying by a common denominator to get rid of the complex fraction, then multiplied by a conjugate to get rid of the square roots. After that I don't know what to do.
April 19th 2010, 05:35 PM
skeeter

Quote:

Originally Posted by soad

So I've been doing good with these until this problem.

Find $f'(a)$

$f(x)=\frac{8}{\sqrt{x+2}}$

This is what we have to start with.

$\frac{\frac{8}{\sqrt{a+h+2}}-\frac{8}{\sqrt{a+2}}}{h}$

I began multiplying by a common denominator to get rid of the complex fraction, then multiplied by a conjugate to get rid of the square roots. After that I don't know what to do.

you should be up to this point ...

$8 \lim_{h \to 0} \frac{1}{h}\left(\frac{-h}{\sqrt{a+h+2} \cdot \sqrt{a+2} \left[\sqrt{a+h+2}+\sqrt{a+2}\right]}\right)$

note that the $h$ 's cancel, allowing the limit to be evaluated as ...

$-\frac{8}{(a+2) \cdot 2\sqrt{a+2}} = -\frac{4}{(a+2)^{\frac{3}{2}}}$
April 19th 2010, 05:59 PM
soad

Okay... so the $h$ 's cancel out because it's approaching $0$ ?

If so, I completely understand now.

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