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

1 Answer
Jan 5, 2016

Answer:

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#