What is the d/(dt)((x)(2x^2+2)) ?

The d/(dt)((x)(2x^2+2)) is 6t^2+2

What is the first d/(dt)((x)(2x^2+2)) ?

The first d/(dt)((x)(2x^2+2)) is 6t^2+2

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) (2 t^{2} + 2))

6 t^{2} + 2

Solution steps

\frac{d}{d t} ((t) (2 t^{2} + 2))

Apply the Product Rule:

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

f = t, g = 2 t^{2} + 2

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

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

\frac{d}{d t} (2 t^{2} + 2) = 4 t

= 1 \cdot (2 t^{2} + 2) + 4 tt

Simplify

1 \cdot (2 t^{2} + 2) + 4 tt : 6 t^{2} + 2

= 6 t^{2} + 2

- \frac{6}{( 2 + x ) ^{4}}

[e^{x + c} \cdot (x + 1)] y

\frac{( 3 - 2 x ) ^{3}}{3}

\frac{2 x}{( ( x - 2 ) ( x + 2 ) )}

\frac{3 ( x + 1 )}{2 x}