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

1 Answer
Jim G.
Dec 27, 2017

#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#

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