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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

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

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

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

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

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

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

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