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second derivative

Posted January 13th, 2008 by Anonymous
in

hey can you help me to find y(x) if

$$y"(x)= \frac{20x}{(x^2-4)^2}$$
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Comments

Antiderivative

On January 14th, 2008 Structure says:
$$y"= \frac{20x}{(x^2-4)^2}$$

Then

$$y'(x)=- \frac{10}{(x^2-4)}$$

and

$$y(x)=-\frac{5}{2}\ln|\frac{x-2}{x+2}|$$

How can you come up with the first step?

On December 14th, 2009 peterszt says:

How can you come up with the first step?

answer

On December 14th, 2009 Structure says:

I know that

$$\frac{1}{x}'=-\frac{1}{x^2}$$

So you have also

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

when I see a fraction

$$\frac{g}{f^2}$$

it worth to verify if g=f '

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