How do you differentiate f(x)=csc(sqrt(e^x)) f(x)=csc(ex) using the chain rule?

1 Answer
Jan 5, 2016

f'(x)=(-sqrt(e^x)csc(sqrt(e^x))cot(sqrt(e^x)))/2

Explanation:

Use the chain rule (a lot of times!):

First issue: the cosecant function: d/dx(csc(u))=-csc(u)cot(u)*u'

f'(x)=-csc(sqrt(e^x))cot(sqrt(e^x))*d/dx(sqrt(e^x))

Second issue: the square root: d/dx(sqrtu)=d/dx(u^(1/2))=1/2u^(-1/2)*u'

f'(x)=-csc(sqrt(e^x))cot(sqrt(e^x))*1/2(e^x)^(-1/2)*d/dx(e^x)

Here, recall that d/dx(e^x)=e^x.

f'(x)=-csc(sqrt(e^x))cot(sqrt(e^x))*1/2(e^x)^(-1/2)*e^x

To multiply (e^x)^(-1/2)*e^x, add the exponents to get (e^x)^(1/2)=sqrt(e^x).

f'(x)=(-sqrt(e^x)csc(sqrt(e^x))cot(sqrt(e^x)))/2