What is the d/(dt)((s)(9+0.02s)) ?

The d/(dt)((s)(9+0.02s)) is 9+0.04t

What is the first d/(dt)((s)(9+0.02s)) ?

The first d/(dt)((s)(9+0.02s)) is 9+0.04t

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) (9 + 0.02 t))

9 + 0.04 t

Solution steps

\frac{d}{d t} ((t) (9 + 0.02 t))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = 9 + 0.02 t

= \frac{d t}{d t} (9 + 0.02 t) + \frac{d}{d t} (9 + 0.02 t) t

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

\frac{d}{d t} (9 + 0.02 t) = 0.02

= 1 \cdot (9 + 0.02 t) + 0.02 t

Simplify

= 9 + 0.04 t

(3 + 2) (3 + 2)

2 + \frac{1}{12} (8.1 - 8) - \frac{\frac{1}{144} ( 8.1 - 8 ) ^{2}}{2}

\frac{3 s + 7}{s ( s + 2 )}

\frac{e ^{- s}}{s ( s - 1 )}

, 6 x^{2} + 17 x + 5 (3 x + a) (b x + c)

What is the d/(dt)((s)(9+0.02s)) ?
The d/(dt)((s)(9+0.02s)) is 9+0.04t
What is the first d/(dt)((s)(9+0.02s)) ?
The first d/(dt)((s)(9+0.02s)) is 9+0.04t