What is the d/(dt)((2t+7)/(2t)) ?

The d/(dt)((2t+7)/(2t)) is -7/(2t^2)

What is the first d/(dt)((2t+7)/(2t)) ?

The first d/(dt)((2t+7)/(2t)) is -7/(2t^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} (\frac{2 t + 7}{2 t})

- \frac{7}{2 t ^{2}}

Solution steps

\frac{d}{d t} (\frac{2 t + 7}{2 t})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d t} (\frac{2 t + 7}{t})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{2} \cdot \frac{\frac{d}{d t} ( 2 t + 7 ) t - \frac{d t}{d t} ( 2 t + 7 )}{t ^{2}}

\frac{d}{d t} (2 t + 7) = 2

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

= \frac{1}{2} \cdot \frac{2 t - 1 \cdot ( 2 t + 7 )}{t ^{2}}

Simplify

\frac{1}{2} \cdot \frac{2 t - 1 \cdot ( 2 t + 7 )}{t ^{2}} : - \frac{7}{2 t ^{2}}

= - \frac{7}{2 t ^{2}}

\frac{d}{d x} (- 2 tan (x))

\int (4 x - 1)^{3} d x

\int_{1}^{6} (2 x^{3} + 5 x^{2} + 2 x + 8) d x

x \to 3 + lim (\frac{1}{6 - 2 x})

f (x) = 1.1 x + e^{x}, at x = 0

What is the d/(dt)((2t+7)/(2t)) ?
The d/(dt)((2t+7)/(2t)) is -7/(2t^2)
What is the first d/(dt)((2t+7)/(2t)) ?
The first d/(dt)((2t+7)/(2t)) is -7/(2t^2)