What is the first d/(dt)(t^2+4t) ?

The first d/(dt)(t^2+4t) is 2t+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

d/(dt)(t^2+4t)

\frac{d}{d t} (t^{2} + 4 t)

2 t + 4

Solution steps

\frac{d}{d t} (t^{2} + 4 t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{2}) + \frac{d}{d t} (4 t)

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

\frac{d}{d t} (4 t) = 4

= 2 t + 4

y^{^{'}} + 10 y = 1

x \to 0 lim (2 + (\frac{3}{x}))

\frac{\partial}{\partial y} (x y + \frac{1}{x} + \frac{1}{y})

\int tan^{3} (10 x) d x

\int \frac{5 x - 3}{x ^{2} - 2 x - 3} d x