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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial x)(x/(1+x^2+y^2))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Quotient Rule:

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

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

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

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

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

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

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

a re a y = 4 x^{2}, y = 4

\int \frac{2 x - 3}{x ^{2} + 6 x + 4} d x

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

(x^{2} + 1) y^{^{'}} + 4 x (y - 1) = 0, y (0) = 3

\frac{d y}{d x} + y^{3} x + 8 y = 0

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