What is the derivative of (4x)/(x+1) ?

The derivative of (4x)/(x+1) is 4/((x+1)^2)

What is the first derivative of (4x)/(x+1) ?

The first derivative of (4x)/(x+1) is 4/((x+1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{4 x}{x + 1}

\frac{4}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x}{x + 1})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{x}{x + 1})

Apply the Quotient Rule:

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

= 4 \cdot \frac{\frac{d x}{d x} ( x + 1 ) - \frac{d}{d x} ( x + 1 ) x}{( x + 1 ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (x + 1) = 1

= 4 \cdot \frac{1 \cdot ( x + 1 ) - 1 \cdot x}{( x + 1 ) ^{2}}

Simplify

4 \cdot \frac{1 \cdot ( x + 1 ) - 1 \cdot x}{( x + 1 ) ^{2}} : \frac{4}{( x + 1 ) ^{2}}

= \frac{4}{( x + 1 ) ^{2}}

\int e^{- 5 x} sin (4 x) d x

\frac{d y}{d x} = \frac{x}{y}, y (0) = - 4

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = 0

ln (x), x = 4

y^{^{''}} + y = sin (x), y (0) = 1, y (\frac{π}{2}) = 0

What is the derivative of (4x)/(x+1) ?
The derivative of (4x)/(x+1) is 4/((x+1)^2)
What is the first derivative of (4x)/(x+1) ?
The first derivative of (4x)/(x+1) is 4/((x+1)^2)