<s>[INST] 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.
 [/INST] 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.
</s>