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

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Answer 1 · 2016-01-05T03:34:22

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

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 $(e^x)^(1/2)=sqrt(e^x)$ .

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

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