What is the derivative of #t^2*2^t#?
1 Answer
Dec 5, 2017
Explanation:
#"differentiate using the "color(blue)"product rule"#
#"given "y=f(t)g(t)" then"#
#dy/dt=f(t)g'(t)+g(t)f'(t)larrcolor(blue)"product rule"#
#f(t)=t^2rArrf'(t)=2t#
#g(t)=2^trArrg'(t)=2^t ln2#
#rArrd/dt(t^2 .2^t)#
#=t^2 2^tln2+2t.2^t#
