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

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Answer 1 · 2015-12-28T21:02:56

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#.

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)#