Question

How do you differentiate `f(x)=cot(sqrt(x^2-1))` using the chain rule?

Answer 1

The explanation is given below.

Explanation:

`f(x) =cot(sqrt(x^2-1))`

Let us first write the same as the following.

`y = cot(u)` and `u=sqrt(v)` and `v=x^2-1`
The above steps are like breaking the chain into manageable links.

Chain rule would work like this:

`dy/dx = dy/(du) * (du)/(dv) * (dv)/(dx)` Chain rule

`y=cot(u)`
`dy/(du) = -sec^2(u)`

`u=sqrt(v)`
`(du)/(dv) = 1/(2sqrt(v))`

`v=x^2-1`
`(dv)/(dx) = 2x`

`f'(x) = dy/dx = -sec^2(u) * 1/(2sqrt(v)) * 2x`
Simplifying

`f'(x) = dy/dx =(- x*sec^2(sqrt(v)))/sqrt(v)`
`f'(x) = dy/dx = (-x*sec^2(sqrt(x^2-1))/sqrt(x^2-1))`

`f'(x) = (-x*sec^2(sqrt(x^2-1))/sqrt(x^2-1))` answer