Question

How do you differentiate `f(x)=cot(1/sqrt(x-3))` using the chain rule?

Answer 1

Applying a "two-degrees" chain rule.

Just remembering the concept: `(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)`

Explanation:

Here, we can name `u=(v)^(-1/2)` and `v=x-3`

Also, we must remember that the rule to derivate `cota` is

`cota=-a'csc^2a`

`(dy)/(dx)=-u'csc^2u*(-1/(2v^(3/2)))*(1)`

`(dy)/(dx)=-(-1/(2v^(3/2)))csc^2(v^(-1/2))*(-1/(2(x-3)^(3/2)))`

`(dy)/(dx)=(csc^2(x-3)^(-1/2))/(2(x-3)^(3/2))(-1/(2(x-3)^(3/2)))`

`(dy)/(dx)=-(csc^2(1/(x-3)^(1/2)))/(4(x-3)^3)`

`(dy)/(dx)=-(csc^2(1/(sqrt(x-3))))/(4(x-3)^3)`