# Math Help - Differentiation help

1. ## Differentiation help

$\displaystyle y=(1-x^2)^3(6+2x)^{-3}$

What I did was:

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

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

$\displaystyle=\frac{(1-x^2)^2(6+2x)^2(6x^2-12x-48)}{6+2x)^6}$

$\displaystyle=\frac{6(1-x^2)^2(x^2-2x-8)}{(6+2x)^4}$

While the answer in the back of the book is:

$\displaystyle=\frac{-6(1-x^2)^2(x^2+6x+1)}{(6+2x)^4}$

Do I have a calculation error or is the way I handled it wrong? Thanks!

2. I think you made a mistake in step 3:

Originally Posted by youngb11
$\displaystyle y=(1-x^2)^3(6+2x)^{-3}$

What I did was:

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

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

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

3. Also, it is very bad practiice to write a series of "=" for things that are not actually equal! The very first line in your chain:
$= \frac{(1- x^2)^3}{(6+ 2x)^3}$
is, in fact equal to y. Everything after that is equal to its derivative. The very next line should be
$y'= \displaystyle=\frac{-6x(6+2x)^3(1-x^2)^2-6(1-x^2)^3(6+2x)^2}{(6+2x)^6}$
and after that it would be alright to just write "= ".

While using the quotient rule works, personally, I would have used the product rule from the original form.

4. Originally Posted by HallsofIvy
While using the quotient rule works, personally, I would have used the product rule from the original form.
Indeed, the preference is almost unavoidable when the denominator is a power of other than one - because then squaring produces a denominator that usually calls for simplifying later, whereas the product rule heads straight for the simpler version.

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside each leg of the legs-uncrossed version of...

... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

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