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

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

Popular >

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

= y \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

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

Simplify

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

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

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

\frac{\partial}{\partial x} (cos (3 x + 4 y))

x \to 1 lim (3 - 2 x)

\int sin (x) \cdot sin (2 x) d x

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