What is the derivative of f(x)=((x^2-4x))/(x+1) ?

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 0 + lim (ln (\frac{7}{x ^{5}}))

15 y^{^{''}} + 7 y^{^{'}} - 2 y = 0

\int sin (π x) cos (π x) d x

\frac{d}{d x} (cosh (a x))

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