What is (\partial)/(\partial x)((xy)/(5x+3y)) ?

The answer to (\partial)/(\partial x)((xy)/(5x+3y)) is (3y^2)/((5x+3y)^2)

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(\partial)/(\partial x)((xy)/(5x+3y))

\frac{\partial}{\partial x} (\frac{x y}{5 x + 3 y})

\frac{3 y ^{2}}{( 5 x + 3 y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{x y}{5 x + 3 y})

Treat

y

as a constant

Take the constant out:

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

= y \frac{\partial}{\partial x} (\frac{x}{5 x + 3 y})

Apply the Quotient Rule:

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

= y \frac{\frac{\partial}{\partial x} ( x ) ( 5 x + 3 y ) - \frac{\partial}{\partial x} ( 5 x + 3 y ) x}{( 5 x + 3 y ) ^{2}}

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

\frac{\partial}{\partial x} (5 x + 3 y) = 5

= y \frac{1 \cdot ( 5 x + 3 y ) - 5 x}{( 5 x + 3 y ) ^{2}}

Simplify

y \frac{1 \cdot ( 5 x + 3 y ) - 5 x}{( 5 x + 3 y ) ^{2}} : \frac{3 y ^{2}}{( 5 x + 3 y ) ^{2}}

= \frac{3 y ^{2}}{( 5 x + 3 y ) ^{2}}

y (x + 9) + y^{^{'}} = 0

\frac{\partial}{\partial y} (y^{2} cos (x y^{2}))

f (x) = x^{3} ln (x)

\int (sin (x) y) d y

\frac{d}{d x} (5 e^{- 3 x})

What is (\partial)/(\partial x)((xy)/(5x+3y)) ?
The answer to (\partial)/(\partial x)((xy)/(5x+3y)) is (3y^2)/((5x+3y)^2)