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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

1 - t^{2}

Solution steps

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

Simplify

t - \frac{1}{3} t^{3} : t - \frac{t ^{3}}{3}

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

Apply the Sum/Difference Rule:

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

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

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

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

= 1 - t^{2}

\frac{d}{d x} ((sech (2 x))^{s i n (3 x)})

\frac{d}{d x} (x \cdot (x - 1))

\int x ln (x + 1) d x

\frac{d}{d x} (ln (\frac{e ^{x} - 1}{e ^{x} + 1}))

x \to 7 lim (\frac{1}{x - 7})