What is the first d/(dt)(t^2+t) ?

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

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d/(dt)(t^2+t)

\frac{d}{d t} (t^{2} + t)

2 t + 1

Solution steps

\frac{d}{d t} (t^{2} + t)

Apply the Sum/Difference Rule:

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

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

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

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

= 2 t + 1

f (x) = \frac{3 cos ( x )}{3 x ^{2} + 3 x - 1}

\frac{d}{d x} (ln (c x))

\frac{d}{d x} (a r (x) c, (ln (x)))

\int x (x^{2} - 2) d x