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

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.


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

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

Tags:

Calculus

Derivatives

Chain Rule

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