Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of  $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\frac{d}{dx} (e^{x ^5} \cdot \sin(x)) = [(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]$$
$$\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}$$
$$\frac{d}{dx} \sqrt{(e^{x^5} \sin(x))} = \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
Therefore, since $\frac{d}{dx} \cos(x) = -\sin(x)$, I have
$$ f'(x) = -\sin(\sqrt{e^{x^5} \sin(x)}) \cdot \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
However, the website I'm using, "WeBWorK", says this is incorrect.
 A: I differentiated your function and did not look at your result. As you can see, they're identical:
$$\begin{align}
\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)\right]
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)'\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left(e^{x^5}\cdot\sin{x}\right)'\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left[\left(e^{x^5}\right)'\cdot\sin{x}+e^{x^5}\cdot\left(\sin{x}\right)'\right]\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left[e^{x^5}\left(x^5\right)'\cdot\sin{x}+e^{x^5}\cdot\cos{x}\right]\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\frac{1}{2\sqrt{e^{x^5}\cdot\sin{x}}}\left[5e^{x^5}x^4\cdot\sin{x}+e^{x^5}\cdot\cos{x}\right]
\end{align}$$
Your differentiation skills are fine. It's the problem with the website that you're suing.
A: This is one case where logarithmic differentiation can make life easier.
$$
\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\,\sin(x)}\right)\right]
=-\sin{\left(\sqrt{e^{x^5}\,\sin(x)}\right)}\,\,\frac{d}{dx}\left[\sqrt{e^{x^5}\,\sin(x)}\right]$$
Let
$$f=\sqrt{e^{x^5}\,\sin(x)}\implies \log(f)=\frac 12 x^5 +\frac 12 \log(\sin(x))$$
$$\frac {f'}f=\frac{1}{2} \left(5 x^4+\cot (x)\right)\implies f'=\frac{1}{2} \left(5 x^4+\cot (x)\right)\sqrt{e^{x^5}\,\sin(x)}$$
$$\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\,\sin(x)}\right)\right]=-\frac{1}{2}\sin{\left(\sqrt{e^{x^5}\,\sin(x)}\right)}\,\, \left(5 x^4+\cot (x)\right)\sqrt{e^{x^5}\,\sin(x)}$$
A: It may be a bit overkill but you can actually integrate your result and show it is equivalent to what you started with
$$ -\frac{1}{2}\int \left(\sin(\sqrt{e^{x^5} \sin(x)} \right) \frac{5x^4e^{x^5}   \sin(x) + \cos(x) e^{x^5}}{\sqrt{e^{x^5} \sin(x)}} dx,$$ 
let 
$$u(x) = \sqrt{e^{x^5} \sin(x)}, \qquad du = \frac{5x^4e^{x^5} \sin(x) + \cos(x) e^{x^5}}{\sqrt{e^{x^5} \sin(x)}} dx, $$
so the integral is simply
$$ - \int \sin(u)du, $$
which is what you started with. 
A: Use the Chain Rule:
$\dfrac{\mathrm d}{\mathrm dx}\cos(\sqrt{e^{x^5}\sin x})$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm dx}(e^{x^5}\sin x)$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \left(\sin x\dfrac{\mathrm d}{\mathrm dx}e^{x^5}+e^{x^5}\dfrac{\mathrm d}{\mathrm dx}\sin x\right)$
$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \left[\sin x\left(\dfrac{\mathrm d}{\mathrm d(x^5)}e^{x^5}\cdot\dfrac{\mathrm d}{\mathrm dx}x^5\right)+e^{x^5}\dfrac{\mathrm d}{\mathrm dx}\sin x\right]$
$=-\sin(\sqrt{e^{x^5}\sin x})\cdot\dfrac{1}{2\sqrt{e^{x^5}\sin x}}\cdot \left[\sin x\cdot  e^{x^5}\cdot 5x^4+e^{x^5}\cdot\cos x\right]$
