# How do you find the derivative of g(t)=e^(-3/t^2)?

Dec 27, 2017

$g ' \left(t\right) = \frac{6 {e}^{- \frac{3}{t} ^ 2}}{{t}^{3}}$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "g(t)=f(h(t))" then}$

$g ' \left(t\right) = f ' \left(h \left(t\right)\right) \times h ' \left(t\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$g \left(t\right) = {e}^{- \frac{3}{t} ^ 2}$

$\Rightarrow g ' \left(t\right) = {e}^{- \frac{3}{t} ^ 2} \times \frac{d}{\mathrm{dt}} \left(- \frac{3}{t} ^ 2\right)$

$\frac{d}{\mathrm{dt}} \left(- \frac{3}{t} ^ 2\right) = \frac{d}{\mathrm{dt}} \left(- 3 {t}^{-} 2\right)$

$= 6 {t}^{-} 3 = \frac{6}{t} ^ 3$

$\Rightarrow g ' \left(t\right) = {e}^{- \frac{3}{t} ^ 2} \times \frac{6}{t} ^ 3$

$\textcolor{w h i t e}{\Rightarrow g ' \left(t\right)} = \frac{6 {e}^{- \frac{3}{t} ^ 2}}{t} ^ 3$