What is the derivative of s(t)=t^3-5t+8 ?

The derivative of s(t)=t^3-5t+8 is 3t^2-5

What is the first derivative of s(t)=t^3-5t+8 ?

The first derivative of s(t)=t^3-5t+8 is 3t^2-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

s (t) = t^{3} - 5 t + 8

3 t^{2} - 5

Solution steps

\frac{d}{d t} (t^{3} - 5 t + 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{3}) - \frac{d}{d t} (5 t) + \frac{d}{d t} (8)

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

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

\frac{d}{d t} (8) = 0

= 3 t^{2} - 5 + 0

Simplify

= 3 t^{2} - 5

y = x, (1, 1)

\frac{x}{x + 3}

n = 1 \sum \infty (\frac{3}{5})^{n}

\frac{d}{d t} (2 sin (t) - sin (2 t))

\int (6 x - 1) d x

What is the derivative of s(t)=t^3-5t+8 ?
The derivative of s(t)=t^3-5t+8 is 3t^2-5
What is the first derivative of s(t)=t^3-5t+8 ?
The first derivative of s(t)=t^3-5t+8 is 3t^2-5