Question

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

Answer 1

Recalling the chain rule, which states that `(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)`, we can rename `u=ln(v)` and `v=1-e^x`.

Explanation:

Doing it step-by-step, separately (which is not necessarilt advised, but this is a quite simple function, so it is ok), we'll have that `f(x)=csc(u)`

`(dy)/(du)=-cscucotu`

`(du)/(dv)=1/v`

`(dv)/(dx)=-e^x`

Aggregating and substituting `u`:

`(dy)/(dx)=-csc(ln(v))cot(lnv)(1/v)(-e^x)`

Substituting `v`:

`(dy)/(dx)=(e^xcsc(ln(1-e^x))cot(ln(1-e^x)))/(1-e^x)`