Question

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

Answer 1

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

Explanation:

`f(x)=sqrt(e^cot(x))`
To find the derivative of f(x), we need to use chain rule .

`color(red)"chain rule :f(g(x))'=f'(g(x)).g'(x)"`
Let `u(x)=cot(x)=>u'(x)=-csc^2(x)`
and `g(x)=e^(x)=>g'(x)=e^(x).g'(u(x))=e^cot(x)`
`f(x)=sqrt(x)=>f'(x)=1/(2sqrt(x))=>f'(g(u(x)))=1/(2sqrt(e^cot(x))`

`d/dx(f(g(u(x)))=f'(g(u(x))).g'(u(x)).u'(x)`
=`1/(sqrt(e^cot(x)))e^cot(x).-cos^2(x)`
=`(-e^cot(x)csc^2x)/sqrt(e^cot(x))`
`color(blue)"cancel the e^cot(x) with sqrt(e^cot(x)) in the denominator"`
=-`(sqrt(e^cot(x)).csc^2(x))/2`