What is the d/(dt)(t^3-5t^2+9t) ?

The d/(dt)(t^3-5t^2+9t) is 3t^2-10t+9

What is the first d/(dt)(t^3-5t^2+9t) ?

The first d/(dt)(t^3-5t^2+9t) is 3t^2-10t+9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (t^{3} - 5 t^{2} + 9 t)

3 t^{2} - 10 t + 9

Solution steps

\frac{d}{d t} (t^{3} - 5 t^{2} + 9 t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{3}) - \frac{d}{d t} (5 t^{2}) + \frac{d}{d t} (9 t)

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

\frac{d}{d t} (5 t^{2}) = 10 t

\frac{d}{d t} (9 t) = 9

= 3 t^{2} - 10 t + 9

y = 2 x + 2, y = x^{2} - 2

y^{2} = 2 x, y = - 4 + x

\int \frac{1}{25 - 4 x ^{2}} d x

\int t \cdot cos (t) d t

\int 18 sin^{3} (x) cos^{2} (x) d x

What is the d/(dt)(t^3-5t^2+9t) ?
The d/(dt)(t^3-5t^2+9t) is 3t^2-10t+9
What is the first d/(dt)(t^3-5t^2+9t) ?
The first d/(dt)(t^3-5t^2+9t) is 3t^2-10t+9