What is the first d/(dt)(t^3-1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

d/(dt)(t^3-1)

\frac{d}{d t} (t^{3} - 1)

3 t^{2}

Solution steps

\frac{d}{d t} (t^{3} - 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{3}) - \frac{d}{d t} (1)

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

\frac{d}{d t} (1) = 0

= 3 t^{2} - 0

Simplify

= 3 t^{2}

\int \frac{1 + x + 1}{1 + x - 1} d x

\int \frac{tan ( x ) + sin ( x )}{sec ( x )} d x

\int 6 x e^{5 x} d x

\int x^{2} (5 x^{3} - 1)^{4} d x

\int \frac{x + 1}{x ^{2} - 16} d x