What is the derivative of (6x)/(x^2+1) ?

The derivative of (6x)/(x^2+1) is (6(-x^2+1))/((x^2+1)^2)

What is the first derivative of (6x)/(x^2+1) ?

The first derivative of (6x)/(x^2+1) is (6(-x^2+1))/((x^2+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{6 x}{x ^{2} + 1}

\frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{x ^{2} + 1})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{x ^{2} + 1})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d x}{d x} ( x ^{2} + 1 ) - \frac{d}{d x} ( x ^{2} + 1 ) x}{( x ^{2} + 1 ) ^{2}}

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

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

= 6 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}} : \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

= \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

\frac{\partial}{\partial y} (y e^{x} + sin (y))

- 7 sin (7 x)

n = 1 \sum \infty \frac{n}{( n + 1 )}

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

x^{2} y^{^{'}} = x y + y^{2}

What is the derivative of (6x)/(x^2+1) ?
The derivative of (6x)/(x^2+1) is (6(-x^2+1))/((x^2+1)^2)
What is the first derivative of (6x)/(x^2+1) ?
The first derivative of (6x)/(x^2+1) is (6(-x^2+1))/((x^2+1)^2)