How do I differentiate $\sin(x)^{\cos(x^3)}$? I'm taking a calc 1 class and I have come across a function that I'm having difficulty finding answers on the web.
$y=\sin(x)^{\cos(x^3)}$
I know there's some chain rule to apply, but what do I do with the cos(x)?
I am assuming this:
$y' = \cos(x^3).(\sin(x)')^{\cos(x^3) - 1}$
Is my thinking correct?
 A: Use that
$$y=e^{\cos (x^3)\log(\sin x)}$$
and by chain rule we have
$$y=e^{f(x)}\implies y'=f'(x)e^{f(x)}$$
A: Another way to solve this is with a differential rule rarely taught: the differential of $q^m$ (I'm using these instead of $u$ and $v$ because the font makes $v$ look too much like $u$).  Then you don't have to do any weird manipulations.  The rule is:
$$
d(q^m) = mq^{m - 1}\,dq + \ln(q)q^m\,dm
$$
Note that this is just a combination of the power rule and the exponential rule.
So, plugging in your values, that would be:
$$q = \sin(x) \\ dq = \cos(x)\,dx$$ and $$m = \cos(x^3) \\ dm = -\sin(x^3) 3x^2\,dx$$
Plugging those into the formula gives:
$$
\cos(x^3)\sin(x)^{\cos(x^3) - 1} \cos(x)\,dx + \ln(\sin(x))\sin(x)^{\cos(x^3)}(-\sin(x^3) 3x^2)\, dx$$
Therefore:
$$\frac{dy}{dx} = \cos(x^3)\sin(x)^{\cos(x^3) - 1} \cos(x) + \ln(\sin(x))\sin(x)^{\cos(x^3)} (-\sin(x^3) 3x^2) $$
It's still ugly, but the nice thing about using the rule is that there's no manipulation - you can basically just apply rules mechanistically (I hope I did all that right - it is sometimes difficult to type in a bunch of different terms and keep them all straight).
