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

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

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

\frac{d}{d 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

\frac{1}{x ^{2}}, at x = 2

\frac{d}{d x} (\frac{cos ( 2 x )}{sin ( 4 x )})

f (x) = x^{3} + x^{2} + x + 5

x \to 0 - lim (\frac{1}{1 - 2 ^{\frac{1}{x}}})

\int 9 x - x d x