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# What is the derivative of this? y = t^(2) √(t-2)

Mar 8, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 t \sqrt{t - 2} + {t}^{2} / \left(2 \sqrt{t - 2}\right)$

#### Explanation:

By product rule, let $f \left(x\right) = {t}^{2}$, let $g \left(x\right) = \sqrt{t - 2}$

$\frac{d}{\mathrm{dx}} f \left(x\right) \cdot g \left(x\right)$ = $f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

$f ' \left(x\right) = 2 t$, $g ' \left(x\right) = \frac{1}{2 \sqrt{t - 2}}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 t \sqrt{t - 2} + {t}^{2} / \left(2 \sqrt{t - 2}\right)$