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

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

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Answer 1 · 2017-02-10T00:46:37

$\frac{d s}{d t} = t^{2} + t + 1$ $(ds)/dt=t^2+t+1$

Explanation:

We will use the rules:

$\frac{d}{d t} [f (t) + g (t)] = \frac{d}{d t} f (t) + \frac{d}{d t} g (t)$ $d/dt[f(t)+g(t)]=d/dtf(t)+d/dtg(t)$
$\frac{d}{d t} t^{n} = n t^{n - 1}$ $d/dtt^n=nt^(n-1)$
$\frac{d}{d t} a \cdot f (t) = a \cdot \frac{d}{d t} f (t)$ $d/dta*f(t)=a*d/dtf(t)$

Then:

$\frac{d s}{d t} = \frac{1}{3} (\frac{d}{d t} t^{3}) + \frac{1}{2} (\frac{d}{d t} t^{2}) + \frac{d}{d t} t^{1}$ $(ds)/dt=1/3(d/dtt^3)+1/2(d/dtt^2)+d/dtt^1$

$\frac{d s}{d t} = \frac{1}{3} (3 t^{2}) + \frac{1}{2} (2 t^{1}) + 1 t^{0}$ $(ds)/dt=1/3(3t^2)+1/2(2t^1)+1t^0$

$\frac{d s}{d t} = t^{2} + t + 1$ $(ds)/dt=t^2+t+1$

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

1 Answer

Explanation:

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