Question

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

Answer 1

`g'(t)=(6e^(-3/t^2))/(t^3)`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "g(t)=f(h(t))" then"`

`g'(t)=f'(h(t))xxh'(t)larrcolor(blue)"chain rule"`

`g(t)=e^(-3/t^2)`

`rArrg'(t)=e^(-3/t^2)xxd/dt(-3/t^2)`

`d/dt(-3/t^2)=d/dt(-3t^-2)`

`=6t^-3=6/t^3`

`rArrg'(t)=e^(-3/t^2)xx6/t^3`

`color(white)(rArrg'(t))=(6e^(-3/t^2))/t^3`