How do you find the derivative of #s=1/3t^3+1/2t^2+t#?
1 Answer
Feb 10, 2017
Explanation:
We will use the rules:
#d/dt[f(t)+g(t)]=d/dtf(t)+d/dtg(t)# #d/dtt^n=nt^(n-1)# #d/dta*f(t)=a*d/dtf(t)#
Then:
#(ds)/dt=1/3(d/dtt^3)+1/2(d/dtt^2)+d/dtt^1#
#(ds)/dt=1/3(3t^2)+1/2(2t^1)+1t^0#
#(ds)/dt=t^2+t+1#
