Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6 - Page 158: 15

$f'(x)=-\frac{2\cos\left(\frac{1}{x^2}\right)}{x^3}$

$f(x)=sin(\frac{1}{x^2})$ $f(x)=sin(x^{-2})$ $f'(x)=cos(x^{-2})\times{-2x^{-2-1}}$ $f'(x)=cos(x^{-2})\times{-2x^{-3}}$ $f'(x)=-\frac{2}{x^3}cos(\frac{1}{x^2})$ $f'(x)=-\frac{2\cos\left(\frac{1}{x^2}\right)}{x^3}$