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

1 Answer
Dec 28, 2015

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)