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I can't seem to figure out how my professor simplified this second derivative. Any help is much appreciated. I'm having trouble simplifying the second derivatives of most problems so step by step instructions would be awesome.

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  • $\begingroup$ From the first line to the second line just factor out $(1-x^2)$ in top and bottom and simplify things in $1-x^2+2(1-x^2)$. $\endgroup$ – Git Gud Mar 23 '14 at 20:44
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Okay. Let me just move some terms around:

$$\frac{(6x)(1-x^2)^2-3(-2x)(1+x^2)[2(1-x^2)]}{(1-x^2)^4}$$

First, you should note that you've got $(1-x^2)$ everywhere. Get rid of it:

$$\frac{(6x)(1-x^2)-3(-2x)(1+x^2)[2]}{(1-x^2)^3}$$

Now note that you can have $6x$ on both terms at the top:

$$\frac{(6x)(1-x^2)+(6x)(1+x^2)[2]}{(1-x^2)^3}$$

(Just multiplied the $-3$ by $-2x$.)

Take the $6x$ out:

$$\frac{(6x)((1-x^2)+(1+x^2)[2])}{(1-x^2)^3}$$

Simplify the inside:

$$\frac{(6x)(1-x^2+2+2x^2)}{(1-x^2)^3}$$

Common term, $x^2$, and some numbers:

$$\frac{(6x)(3+x^2)}{(1-x^2)^3}$$

And there it is! The answer.

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Notice that each term in the numerator has a common factor of $1-x^2$ so cancelling that you are left with

$$\begin{align}\frac{(6x)(1-x^2)^2-3(1+x^2)(2)(1-x^2)(-2x)}{(1-x^2)^4} &= \frac{6x(1-x^2)-3(1+x^2)(2)(-2x)}{(1-x^2)^3} \\ &= \frac{6x(1-x^2)+12x(1+x^2)}{(1-x^2)^3} \\ &= \frac{6x-6x^3+12x+12x^3}{(1-x^2)^3} \\ &= \frac{18x+6x^3}{(1-x^2)^3} \\ &= \frac{6x(3+x^2)}{(1-x^2)^3}\end{align}$$

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