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

1 Answer
Jan 5, 2016

Answer:

#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#