Question

How do you differentiate `f(x)=cot(4x-x^2)` using the chain rule?

Answer 1

`frac{d}{dx}(cot(4x-x^2))=-frac{-2x+4}{sin ^2(4x-x^2)}`

Explanation:

`frac{d}{dx}(cot(4x-x^2))`
Applying chain rule,`frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}`
Let, `4x-x^2 =u`
`=frac{d}{du}(cot (u))frac{d}{dx}(4x-x^2)`

We know,
`frac{d}{du}(cot (u))=-frac{1}{sin ^2(u)}`
and,
`frac{d}{dx}(4x-x^2)=4-2x`

So,
`=(-frac{1}{sin ^2(u)})(4-2x)`

Substituting back,`4x-x^2 =u`

`=(-frac{1}{sin ^2(4x-x^2)})(4-2x)`

Simplifying it,
`-frac{-2x+4}{sin ^2(4x-x^2)}`