What is (\partial)/(\partial y)(((x+y))/((y+z))) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(((x+y))/((y+z)))

\frac{\partial}{\partial y} (\frac{( x + y )}{( y + z )})

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

Solution steps

\frac{\partial}{\partial y} (\frac{( x + y )}{( y + z )})

Treat

x, z

as constants

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (\frac{x}{x ^{6} - y ^{3}})

\frac{d y}{d x} = 4 x + y, y (0) = 9

\frac{d}{d x} (x^{16})

y^{^{''}} - 2 y^{^{'}} = sin (x)

\int x^{2} (2 x - 1)^{5} d x

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