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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (t) = \frac{3}{t}

\frac{d}{d x} (- \frac{3}{4} e^{- \frac{x}{4}})

\frac{\partial}{\partial x} (- e^{- x} cos (y))

\int \frac{4 x ^{5}}{9 - 16 x ^{6}} d x

\frac{d}{d x} ((x^{x} + 1)^{2})

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