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If #f(x) =cot(-x/3) # and #g(x) = sqrt(x^2+1 #, what is #f'(g(x)) #?

1 Answer

Shwetank Mauria

Oct 18, 2016

#f'(g(x))=1/3csc^2(-sqrt(x^2+1)/3)#

Explanation:

As #f(x)=cot(-x/3)#

#f'(x)=d/(d(-x/3))(cot(-x/3))xxd/(dx)(-x/3)#

= #-csc^2(-x/3)xx(-1/3)=1/3csc^2(-x/3)#

As #g(x)=sqrt(x^2+1)#

#f'(g(x))=1/3csc^2(-sqrt(x^2+1)/3)#

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