What is the first d/(dt)(3t-t^3) ?

The first d/(dt)(3t-t^3) is 3-3t^2

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d/(dt)(3t-t^3)

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

3 - 3 t^{2}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 3 - 3 t^{2}

\frac{d}{d x} (x^{2} x)

(3.2) (1.6)

\frac{d}{d x} (2 x^{2} cos (3 x))

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\int 1 + cos (8 x) d x