What is the derivative of $\displaystyle sin^\frac {1}{3} x $?
Actually part of a bigger problem, but I think I can tackle it if I just know this.
let $\displaystyle t=y^{\frac{1}{3}}$ and $\displaystyle y=\sin x $ we need dt/dx
$\displaystyle dt/dx = dt/dy \cdot dy/dx $
$\displaystyle dt/dx = 1/3y^{\frac{-2}{3}} \cdot \cos x $ sub y value
$\displaystyle dt/dx = 1/3(\sin x)^{\frac{-2}{3}}(\cos x )$
$\displaystyle dt/dx = \frac{\cos x }{3(\sin ^{\frac{2}{3}})}$
that's it