What is the derivative of (8x^2+9/(x^2+1)) ?

The derivative of (8x^2+9/(x^2+1)) is -(2x)/((x^2+1)^2)

What is the first derivative of (8x^2+9/(x^2+1)) ?

The first derivative of (8x^2+9/(x^2+1)) is -(2x)/((x^2+1)^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{8 x ^{2} + 9}{x ^{2} + 1})

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

Solution steps

\frac{d}{d x} (\frac{8 x ^{2} + 9}{x ^{2} + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 8 x ^{2} + 9 ) ( x ^{2} + 1 ) - \frac{d}{d x} ( x ^{2} + 1 ) ( 8 x ^{2} + 9 )}{( x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (8 x^{2} + 9) = 16 x

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

= \frac{16 x ( x ^{2} + 1 ) - 2 x ( 8 x ^{2} + 9 )}{( x ^{2} + 1 ) ^{2}}

Simplify

\frac{16 x ( x ^{2} + 1 ) - 2 x ( 8 x ^{2} + 9 )}{( x ^{2} + 1 ) ^{2}} : - \frac{2 x}{( x ^{2} + 1 ) ^{2}}

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

\frac{\partial}{\partial y} (\frac{1}{2} y)

\int - 3 tan^{3} (x) d x

6 y^{^{''}} + 11 y^{^{'}} - 2 y = 0

\int x^{2} 6^{2} - x^{2} d x

\int (x + 3) (x - 4) d x

What is the derivative of (8x^2+9/(x^2+1)) ?
The derivative of (8x^2+9/(x^2+1)) is -(2x)/((x^2+1)^2)
What is the first derivative of (8x^2+9/(x^2+1)) ?
The first derivative of (8x^2+9/(x^2+1)) is -(2x)/((x^2+1)^2)