find dy/dx

**bigwave** · March 5th 2010, 02:18 PM

find $\frac{dy}{dx}$
$ y=x^3(5x^2+1)^{-\frac{2}{3}} $

the answer in book is:
$ \frac{1}{3}x^2(5x^2+1)^{-\frac{5}{3}}(25x^2+9) $

I tried using the product and power rule on this
but did not see how they got the $25x^2+9$
also, this is under the section for implicit differentiation but I only see the variable x in it.

**Archie Meade** · March 5th 2010, 02:39 PM

hi bigwave,

$f'(x)=\left(5x^2+1\right)^{-\frac{3}{2}}3x^2+x^3\left(-\frac{2}{3}\right)\left(5x^2+1\right)^{-\frac{5}{2}}10x$

$=\left(5x^2+1\right)^{-\frac{5}{2}}\left(5x^2+1\right)3x^2-\frac{20x^4}{3}\left(5x^2+1\right)^{-\frac{5}{2}}$

$=\left(5x^2+1\right)^{-\frac{5}{2}}\left(15x^4+3x^2-\frac{20}{3}x^4\right)$

$=\left(5x^2+1\right)^{-\frac{5}{2}}\left(\frac{25}{3}x^4+3x^2\right)$

$=\left(5x^2+1\right)^{-\frac{5}{2}}\left(\frac{x^2}{3}\right)\left(25x^2+ 9\right)$

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