How do I differentiate $\sin(x)^{\cos(x^3)}$?

Question

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?

Answer 1

Use that

$$y=e^{\cos (x^3)\log(\sin x)}$$

and by chain rule we have

$$y=e^{f(x)}\implies y'=f'(x)e^{f(x)}$$

Answer 2

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).