What is the first d/(dt)(t^2+5t) ?

The first d/(dt)(t^2+5t) is 2t+5

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d/(dt)(t^2+5t)

\frac{d}{d t} (t^{2} + 5 t)

2 t + 5

Solution steps

\frac{d}{d t} (t^{2} + 5 t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{2}) + \frac{d}{d t} (5 t)

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

\frac{d}{d t} (5 t) = 5

= 2 t + 5

\frac{\partial}{\partial t} (ln (k x - m t))

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

\frac{d}{d x} (\frac{4 x}{9 y})

\frac{d}{d x} (- 7 (3 x^{4} + 2)^{- 4})

\frac{d}{d x} (\frac{2}{x ^{2} + 5})