What is the derivative of-2t^3+96t ?

The derivative of-2t^3+96t is -6t^2+96

What is the first derivative of-2t^3+96t ?

The first derivative of-2t^3+96t is -6t^2+96

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derivative of

- 2 t^{3} + 96 t

- 6 t^{2} + 96

Solution steps

\frac{d}{d t} (- 2 t^{3} + 96 t)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d t} (2 t^{3}) + \frac{d}{d t} (96 t)

\frac{d}{d t} (2 t^{3}) = 6 t^{2}

\frac{d}{d t} (96 t) = 96

= - 6 t^{2} + 96

\frac{d}{d x} (9 cos (x) - 7 x^{2})

\int \frac{7 x ^{2}}{( x ^{2} + 1 ) ( x + 1 )} d x

x \to 0 lim (\frac{1 - e ^{- x}}{sin ( x )})

x (x^{2} + 1) d y = 1 - y^{2} d x

What is the derivative of-2t^3+96t ?
The derivative of-2t^3+96t is -6t^2+96
What is the first derivative of-2t^3+96t ?
The first derivative of-2t^3+96t is -6t^2+96