Question

How do you differentiate `f(x)=csc(ln(1/x^4))` using the chain rule?

Answer 1

`(4csc(4lnx)cot(4lnx))/x`

Explanation:

This can be rewritten as:
`-csc(4lnx)`

The derivative of:
`csc(u)=-csc(u)cot(u)`
`ln u= 1/u`

Solving by the chain rule:
`d/dx(-csc(4lnx))=csc(4lnx)cot(4lnx)(d/dx[4lnx])`
`=csc(4lnx)cot(4lnx)(4/x)`
`=(4csc(4lnx)cot(4lnx))/x`