How do you find the derivative of $s=1/3t^3+1/2t^2+t$ ?

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Answer 1 · 2017-02-10T00:46:37

$(ds)/dt=t^2+t+1$

We will use the rules:

Then:

$(ds)/dt=1/3(d/dtt^3)+1/2(d/dtt^2)+d/dtt^1$

$(ds)/dt=1/3(3t^2)+1/2(2t^1)+1t^0$

$(ds)/dt=t^2+t+1$

How do you find the derivative of s=1/3t^3+1/2t^2+ts=1/3t^3+1/2t^2+t?