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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

(0, 2), (4, - 2)

\frac{x}{x ^{2} - 6}

\int \frac{1}{16 sin ^{2} ( x )} d x

81 y^{^{''}} + 18 y^{^{'}} + y = 0

(1.2) (2.4)

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