This thing is a monster:

Find $\displaystyle dy/dx$ for $\displaystyle y=\frac{sin^3(5x)\left(4x-x^2\right)^\frac{1}{3}}{1-tan(x)}$

Is there some kind of easy way to differentiate this thing? Am I missing something?

So far I have $\displaystyle y=(st)\left(u\right)^\left(-1\right)$, where $\displaystyle s=sin^3(5x), t=\left(4x-x^2\right)^\frac{1}{3}, u=1-tan(x)$

Product rule, chain rule, blah blah ... I get: $\displaystyle y'=(st)\left(\left(u\right)^\left(-1\right)\right)^'$$\displaystyle +\left(st\right)^'$$\displaystyle \left(u\right)^\left(-1\right)$$\displaystyle =-(st)\left(u\right)^\left(-2\right)$$\displaystyle \left(u\right)^'\right)\right)+\left(st'+ts'\right )\left(u\right)^\left(-1\right)$ Factoring: $\displaystyle y'=\left(u\right)^\left(-1\right)$$\displaystyle \right)\left(st'+ts'-stu'\left(u\right)^\left(-1\right)\right)$

My individual derivatives are $\displaystyle s'=15sin^2(5x)cos(5x), t'=\frac{4-2x}{3\left(4-x^2\right)^\frac{2}{3}}, u'=-sec^2(x)$

SOOOOO putting it all together I get:$\displaystyle y'=-\frac{\left(\frac{sin^3(5x)\left(4-2x\right)}{3\left(4-x^2\right)^\frac{2}{3}}+15sin^2(5x)cos(5x)+\frac{s in^3(5x)\left(4x-x^2\right)^\frac{1}{3}sec^2(x)}{1-tan(x)}\right)}{1-tan(x)}$

And I can factor futher:

$\displaystyle y'=-\frac{sin^2(5x)\left(4x-x^2\right)^\frac{1}{3}\left(\frac{sin(5x)\left(4-2x\right)}{12-3x^2}+15cos(5x)+\frac{sin(5x)sec^2(x)}{1-tan(x)}\right)}{1-tan(x)}$

Does this seem right? Please help I have very little time to submit this and my teacher is a ... ah ... hard-case. She'll mark me down for skipping steps, despite the right answer. Can this be futher simplified? Was there an easier way to go about this?

Thanks for any help you guys can provide!