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

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

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(\partial)/(\partial x)((x^2)/(x+y))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Quotient Rule:

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

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

\frac{\partial}{\partial x} (x^{2}) = 2 x

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

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

Simplify

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

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

x \cdot ln (x)

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

\int \frac{2 x ^{2} - 4}{x ^{3}} d x

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{17}{4}

\frac{d}{d x} (x^{\frac{3}{2}} + 4)

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