# What is the derivative of $f(x)=(\sin^{3}(5x))^{\frac{1}{4}}$?

What is the derivative of the following function?

$$f(x)=(\sin^{3}(5x))^{\frac{1}{4}}$$

So I did the chain rule and I got

$$(\frac{1}{4})((\sin^{3}(5x))^{-3/4})(3\sin^2(5x))(\cos(5x))(5).$$

Does that look right? How do I simplify that? Thank you for the help and feedback!

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Yes..it is correct! – Tapu Feb 5 '13 at 18:41
awesome! how do I simplify that? That's where I seem to get stuck on these sorts of problems. – user56852 Feb 5 '13 at 18:42
$\sin^3(x)=(\sin x)^3$... – Jp McCarthy Feb 5 '13 at 18:43
collect the exponents. will be some multiple of $\frac{\cos(5x)}{\sin^?(5x)}$ – example Feb 5 '13 at 18:43
May be you first simplify your given function as $(\sin 5x)^\frac{3}{4}$..? – Tapu Feb 5 '13 at 18:44

As I said in the comment, you can do better by starting with $$(\sin 5x)^\frac{3}{4}$$
Then, its derivative w.r.t. $x$ is $$\frac{3}{4}(\sin 5x)^{-\frac{1}{4}}.\cos 5x. 5=\frac{15}{4}(\sin 5x)^{-\frac{1}{4}}.\cos 5x$$ Nothing simplifies more!
$\large{\frac{15}{2}(\sin 5x)^{-\frac{5}{4}}.\sin 10x}$ if you apply double angle rule...tops – bryansis2010 Feb 5 '13 at 18:58
applied $\large{\sin 2x = 2(\sin x)(\cos x)}$ – bryansis2010 Feb 5 '13 at 19:03