How to differentiate $\frac{2x^5}{\tan x}$ $$\frac{2x^5}{\tan x}$$
I can differentiate $2x^5$ ($10x^4$) and $\tan x$ ($\sec^2 x$) but can't do that one
Is there a rule I can apply?
 A: Hint: Use the quotient rule which is a little similar to the chain rule, which says that: $$\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{f(x)}{g(x)}\right)  =  \dfrac{g(x) f'(x)-f(x) g'(x)}{g(x)^2}.$$
In your example, let $f(x)=2x^5$, $f'(x)=10x^4$, $g(x)=\tan x$ and finally $g'(x)=\sec^2x$.
I hope this helps. 
Best wishes, $\mathcal H$akim.
A: You have three choices. 
You can keep the function as a fraction, and use the quotient rule:
$$\frac{2x^5}{\tan x}$$
You can write the function as a product, and use the product rule:
$$\frac{2x^5}{\tan x} \equiv 2x^5 \cdot \cot x$$
You can write the function as a product in a different way and use the product and chain rules:
$$\frac{2x^5}{\tan x} \equiv 2x^5 \cdot (\tan x)^{-1}$$
A: Write it as $2x^5\cot x$
You can use product rule : $\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$
A: How to differentiate $2x^5/\tan x$
$$\begin{align}
\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac uv\right) & = \dfrac{v\dfrac{\mathrm du}{\mathrm dx} - u\dfrac{\mathrm dv}{\mathrm dx}}{v^2} \\
\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{2x^5}{\tan x}\right)& = \tan x\cdot10x^4 - 2x^5\cdot(\sec x)^2)/(\tan x)^2 \\
             & =  \tan x\cdot10x^4-2x^5.(1+(\tan x)^2))/(\tan x)^2\\
             & = 10x^4/\tan x - 2x^5.\sec2x \\
             & = 10x^4*\cot x - 2x^5\sec2x \\
\end{align}$$
