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

$f ' \left(x\right) = \frac{1 + {\tan}^{2} \left(\sqrt{\frac{1}{x} ^ 2}\right)}{{x}^{3} \cdot \sqrt{\frac{1}{x} ^ 2}}$
Note that $\left(\tan \left(x\right)\right) ' = 1 + {\tan}^{2} \left(x\right)$
$f ' \left(x\right) = - \left(1 + {\tan}^{2} \left(\sqrt{\frac{1}{x} ^ 2}\right)\right) \cdot \frac{1}{2} \cdot {\left(\frac{1}{x} ^ 2\right)}^{- \frac{1}{2}} \cdot \left(- \frac{2}{x} ^ 3\right)$