Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$

Question

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.

Answer 1

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.

Answer 2

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 $$f=\sqrt{e^{x^5}\,\sin(x)}\implies \log(f)=\frac 12 x^5 +\frac 12 \log(\sin(x))$$ $$\frac {f'}f=\frac{1}{2} \left(5 x^4+\cot (x)\right)\implies f'=\frac{1}{2} \left(5 x^4+\cot (x)\right)\sqrt{e^{x^5}\,\sin(x)}$$

$$\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\,\sin(x)}\right)\right]=-\frac{1}{2}\sin{\left(\sqrt{e^{x^5}\,\sin(x)}\right)}\,\, \left(5 x^4+\cot (x)\right)\sqrt{e^{x^5}\,\sin(x)}$$

Answer 3

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.

Answer 4

Use the Chain Rule:

$\dfrac{\mathrm d}{\mathrm dx}\cos(\sqrt{e^{x^5}\sin x})$

$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm dx}(e^{x^5}\sin x)$

$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \left(\sin x\dfrac{\mathrm d}{\mathrm dx}e^{x^5}+e^{x^5}\dfrac{\mathrm d}{\mathrm dx}\sin x\right)$

$=\dfrac{\mathrm d}{\mathrm d(\sqrt{e^{x^5}\sin x})}\cos(\sqrt{e^{x^5}\sin x})\cdot \dfrac{\mathrm d}{\mathrm d(e^{x^5}\sin x)}(\sqrt{e^{x^5}\sin x})\cdot \left[\sin x\left(\dfrac{\mathrm d}{\mathrm d(x^5)}e^{x^5}\cdot\dfrac{\mathrm d}{\mathrm dx}x^5\right)+e^{x^5}\dfrac{\mathrm d}{\mathrm dx}\sin x\right]$

$=-\sin(\sqrt{e^{x^5}\sin x})\cdot\dfrac{1}{2\sqrt{e^{x^5}\sin x}}\cdot \left[\sin x\cdot e^{x^5}\cdot 5x^4+e^{x^5}\cdot\cos x\right]$