What is the derivative of (x^2+5x+6/(x^2-4)) ?

The derivative of (x^2+5x+6/(x^2-4)) is -5/((x-2)^2)

What is the first derivative of (x^2+5x+6/(x^2-4)) ?

The first derivative of (x^2+5x+6/(x^2-4)) is -5/((x-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 x} (\frac{x ^{2} + 5 x + 6}{x ^{2} - 4})

- \frac{5}{( x - 2 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 5 x + 6}{x ^{2} - 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 5 x + 6 ) ( x ^{2} - 4 ) - \frac{d}{d x} ( x ^{2} - 4 ) ( x ^{2} + 5 x + 6 )}{( x ^{2} - 4 ) ^{2}}

\frac{d}{d x} (x^{2} + 5 x + 6) = 2 x + 5

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

= \frac{( 2 x + 5 ) ( x ^{2} - 4 ) - 2 x ( x ^{2} + 5 x + 6 )}{( x ^{2} - 4 ) ^{2}}

Simplify

\frac{( 2 x + 5 ) ( x ^{2} - 4 ) - 2 x ( x ^{2} + 5 x + 6 )}{( x ^{2} - 4 ) ^{2}} : - \frac{5}{( x - 2 ) ^{2}}

= - \frac{5}{( x - 2 ) ^{2}}

\frac{\partial}{\partial y} (\frac{x ^{2}}{( x - y ) ^{2}})

\int - \frac{1}{x} d x

\frac{d}{d x} (ln (x^{13}))

\frac{d}{d x} (e^{(- 2 x)})

\int 3 y^{2} (e^{y^{3}} - 1) d y

What is the derivative of (x^2+5x+6/(x^2-4)) ?
The derivative of (x^2+5x+6/(x^2-4)) is -5/((x-2)^2)
What is the first derivative of (x^2+5x+6/(x^2-4)) ?
The first derivative of (x^2+5x+6/(x^2-4)) is -5/((x-2)^2)