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

1 Answer
Jan 5, 2016

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