What is the derivative of (t^3)/3+(t^2)/2+t-1 ?

The derivative of (t^3)/3+(t^2)/2+t-1 is t^2+t+1

What is the first derivative of (t^3)/3+(t^2)/2+t-1 ?

The first derivative of (t^3)/3+(t^2)/2+t-1 is t^2+t+1

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derivative of

\frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t - 1

t^{2} + t + 1

Solution steps

\frac{d}{d t} (\frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t - 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (\frac{t ^{3}}{3}) + \frac{d}{d t} (\frac{t ^{2}}{2}) + \frac{d t}{d t} - \frac{d}{d t} (1)

\frac{d}{d t} (\frac{t ^{3}}{3}) = t^{2}

\frac{d}{d t} (\frac{t ^{2}}{2}) = t

\frac{d t}{d t} = 1

\frac{d}{d t} (1) = 0

= t^{2} + t + 1 - 0

Simplify

= t^{2} + t + 1

What is the derivative of (t^3)/3+(t^2)/2+t-1 ?
The derivative of (t^3)/3+(t^2)/2+t-1 is t^2+t+1
What is the first derivative of (t^3)/3+(t^2)/2+t-1 ?
The first derivative of (t^3)/3+(t^2)/2+t-1 is t^2+t+1