How do you find the derivative of #f(x)=sin (1/x^2)# using the chain rule?

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Answer 1 · 2018-04-01T19:02:37

#f^'(x)=-2/x^3cos(1/x^2)#

Here,

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

Using Chain Rule,we get

#f^'(x)=cos(x^-2)d/(dx)(x^-2)#

#=cos(1/x^2)(-2x^(-2-1))#

#=-2cos(1/x^2)(x^-3)#

#=-2/x^3cos(1/x^2)#