# What is the second derivative of f(x)=sin(1/x^2) ?

$f ' \left(x\right) = \cos \left(\frac{1}{x}\right) \cdot - \frac{1}{x} ^ 2$
$f = \cos \left(\frac{1}{x}\right) , g = - \frac{1}{x} ^ 2 \to f ' = - \sin \left(\frac{1}{x}\right) \left(- \frac{1}{x} ^ 2\right) , g ' = \frac{2}{x} ^ 3$
$f ' ' \left(x\right) = f g ' + g f ' = \cos \left(\frac{1}{x}\right) \left(\frac{2}{x} ^ 3\right) + \left(- \frac{1}{x} ^ 2\right) \sin \left(\frac{1}{x}\right) \left(\frac{1}{x} ^ 2\right) = \frac{2}{x} ^ 3 \cos \left(\frac{1}{x} ^ 2\right) + \left(- \frac{1}{x} ^ 4\right) \sin \left(\frac{1}{x}\right)$