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)

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(\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}}

\int 2 sin (2 x) d x

\int (1 - x) x d x

e^{- \frac{x}{2}} e^{- \frac{1}{4}}

\frac{\partial}{\partial x} (6 y cos (x))

x \to 7 lim (\frac{1}{( x - 7 )})

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)