# How do you differentiate sqrtt*(1-t^2)?

It's quickest to multiply $\setminus \sqrt{t}$ through with the distributive property and then use linearity and the power rule:
$\frac{d}{\mathrm{dt}} \left(\setminus \sqrt{t} \setminus \cdot \left(1 - {t}^{2}\right)\right) = \frac{d}{\mathrm{dt}} \left({t}^{\frac{1}{2}} - {t}^{\frac{5}{2}}\right) = \setminus \frac{1}{2} {t}^{- \frac{1}{2}} - \setminus \frac{5}{2} {t}^{\frac{3}{2}} .$
$\frac{d}{\mathrm{dt}} \left(\setminus \sqrt{t} \setminus \cdot \left(1 - {t}^{2}\right)\right) = \setminus \frac{1}{2} {t}^{- \frac{1}{2}} \left(1 - {t}^{2}\right) + {t}^{\frac{1}{2}} \setminus \cdot \left(- 2 t\right)$
$= \setminus \frac{1}{2} {t}^{- \frac{1}{2}} - \setminus \frac{1}{2} {t}^{\frac{3}{2}} - 2 {t}^{\frac{3}{2}} = \setminus \frac{1}{2} {t}^{- \frac{1}{2}} - \setminus \frac{5}{2} {t}^{\frac{3}{2}} .$