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

The answer to (\partial)/(\partial x)((x-2y)/(x^2-y)) is (-x^2+4xy-y)/((x^2-y)^2)

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

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

\frac{- x ^{2} + 4 x y - y}{( x ^{2} - y ) ^{2}}

Solution steps

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

\frac{\partial}{\partial x} (x - 2 y) = 1

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

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

Simplify

\frac{1 \cdot ( x ^{2} - y ) - 2 x ( x - 2 y )}{( x ^{2} - y ) ^{2}} : \frac{- x ^{2} + 4 x y - y}{( x ^{2} - y ) ^{2}}

= \frac{- x ^{2} + 4 x y - y}{( x ^{2} - y ) ^{2}}

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

\frac{d}{d x} (cos (15 x))

\int arctan (6 x) d x

y^{3} - (10 x + 6) + 3 x y^{2} y^{^{'}} = 0

\int \frac{sin ^{3} ( x )}{cos ^{4} ( x )} d x

What is (\partial)/(\partial x)((x-2y)/(x^2-y)) ?
The answer to (\partial)/(\partial x)((x-2y)/(x^2-y)) is (-x^2+4xy-y)/((x^2-y)^2)