What is the derivative of g(t)=t^3+3t ?

The derivative of g(t)=t^3+3t is 3t^2+3

What is the first derivative of g(t)=t^3+3t ?

The first derivative of g(t)=t^3+3t is 3t^2+3

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derivative of

g (t) = t^{3} + 3 t

3 t^{2} + 3

Solution steps

\frac{d}{d t} (t^{3} + 3 t)

Apply the Sum/Difference Rule:

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

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

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

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

= 3 t^{2} + 3

5 e^{x} cos (x)

\frac{d}{d x} (arctan (\frac{300}{x}))

h (x) = ln (x + x^{2} - 6)

\int sec^{6} (6 x) d x

\int e^{x} 4 x d x

What is the derivative of g(t)=t^3+3t ?
The derivative of g(t)=t^3+3t is 3t^2+3
What is the first derivative of g(t)=t^3+3t ?
The first derivative of g(t)=t^3+3t is 3t^2+3