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

Mar 5, 2017

$\frac{d}{\mathrm{dx}} \sqrt{\frac{1}{x} ^ 2} = - \frac{\left\mid x \right\mid}{x} ^ 3$

#### Explanation:

You can name:

$y \left(x\right) = \frac{1}{x} ^ 2 = {x}^{-} 2$

so that:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \left(\sqrt{y}\right) \frac{d}{\mathrm{dx}} \left({x}^{-} 2\right) = \frac{1}{2 \sqrt{y}} \left(- 2 {x}^{-} 3\right) = \frac{1}{2 \sqrt{\frac{1}{x} ^ 2}} \left(- 2 {x}^{-} 3\right) = - \frac{\sqrt{{x}^{2}}}{x} ^ 3 = - \frac{\left\mid x \right\mid}{x} ^ 3$

You can also note that:

$f \left(x\right) = \sqrt{\frac{1}{x} ^ 2} = \frac{1}{\left\mid x \right\mid}$

so that:

$\left\{\begin{matrix}\frac{\mathrm{df}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right) = - \frac{1}{x} ^ 2 \text{ for " x > 0 \\ (df)/dx = d/dx (-1/x) = 1/x^2" for } x < 0\end{matrix}\right.$

which is clearly the same.