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

The first d/(dt)(t^3+t) is 3t^2+1

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

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

3 t^{2} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 3 t^{2} + 1

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

f (x) = \frac{1}{x + 6}, at x = 3

x^{3}, 2 x - 15.8 x^{2}

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

x - 2, 14 - x, x = 2