What is (\partial)/(\partial x)((-8x)/(x^2+1)) ?

The answer to (\partial)/(\partial x)((-8x)/(x^2+1)) is -(8(-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 >

(\partial)/(\partial x)((-8x)/(x^2+1))

\frac{\partial}{\partial x} (\frac{- 8 x}{x ^{2} + 1})

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

Solution steps

\frac{\partial}{\partial x} (\frac{- 8 x}{x ^{2} + 1})

Simplify

\frac{- 8 x}{x ^{2} + 1} : - \frac{8 x}{x ^{2} + 1}

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

= - 8 \cdot \frac{\frac{\partial}{\partial x} ( x ) ( x ^{2} + 1 ) - \frac{\partial}{\partial x} ( x ^{2} + 1 ) x}{( x ^{2} + 1 ) ^{2}}

\frac{\partial}{\partial x} (x) = 1

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

= - 8 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}}

Simplify

- 8 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}} : - \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

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

y = 4 + 5 x^{2} - 2 x^{3}, at x = a

\int \frac{( x ^{3} - 12 x + 4 )}{( x ^{2} - 3 x + 2 ) ^{2}} d x

y^{^{'}} = 5 y + y^{2}

\frac{d}{d x} (x^{2} cos (x^{3}))

(1.4) (2.2)

What is (\partial)/(\partial x)((-8x)/(x^2+1)) ?
The answer to (\partial)/(\partial x)((-8x)/(x^2+1)) is -(8(-x^2+1))/((x^2+1)^2)