How do you differentiate #f(x) = 1/sqrt(sin^2(2-x^2) # using the chain rule?

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Feb 3, 2017

#f(x)=1/(sqrt(sin^2(2-x^2)))=(sin^2(2-x^2))^(-1/2)=((sin(2-x^2))^2)^(-1/2)=(sin(2-x^2))^-1#

In order to differentiate this, we need to use the chain rule twice on #sin(f(x))^n#.

#d/dx[sin(f(x))]^n=n[sin(f(x))]^(n-1)cos(f(x)) f'(x)#

#f'(x)=-1(sin^2(2-x^2))^(-2)cos(2-x^2)-2x=#

#=(2xcos(2-x^2))/(sin^2(2-x^2))^2#

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