What is the d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) ?

The d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) is 1

What is the first d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) ?

The first d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) is 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (\frac{t ^{3} - 2 t ^{2} + 2 t - 4}{t ^{2} + 2})

1

Solution steps

\frac{d}{d t} (\frac{t ^{3} - 2 t ^{2} + 2 t - 4}{t ^{2} + 2})

Simplify

\frac{t ^{3} - 2 t ^{2} + 2 t - 4}{t ^{2} + 2} : - 2 + t

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d t} (2) = 0

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

= - 0 + 1

Simplify

= 1

sin (θ)

\int \frac{1}{x ^{2} - 10 x} d x

y^{^{'}} = y^{2} + 3

f (x) = 2 x^{2} + 3 x - 2, at a = - 1

\int_{5}^{5.89} 0.75 (y - 5)^{2} + 5 d y

What is the d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) ?
The d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) is 1
What is the first d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) ?
The first d/(dt)((t^3-2t^2+2t-4)/(t^2+2)) is 1