How do you find the derivative of #g(t)=e^(-3/t^2)#?
1 Answer
Dec 27, 2017
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#