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

We can rename $u = {t}^{- \frac{1}{2}}$ and proceed, following chain rule statement: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
now, we have $f \left(t\right) = \frac{1}{u} = {u}^{-} 1$
$\frac{\mathrm{df} \left(t\right)}{\mathrm{dt}} = - \frac{1}{u} ^ 2 \left(- \frac{1}{2 {t}^{\frac{3}{2}}}\right) = \frac{1}{\left({\left({t}^{- \frac{1}{2}}\right)}^{2}\right) \left(2 {t}^{\frac{3}{2}}\right)} = \frac{1}{{t}^{- 1} \cdot 2 {t}^{\frac{3}{2}}}$
$\frac{\mathrm{df} \left(t\right)}{\mathrm{dt}} = \frac{1}{2 {t}^{- 1 + \frac{3}{2}}} = \frac{1}{2 {t}^{\frac{1}{2}}}$