What is the d/(dy)((2x^2-xy+2y^2)/(x-y)) ?

The d/(dy)((2x^2-xy+2y^2)/(x-y)) is (x^2+4xy-2y^2)/((x-y)^2)

What is the first d/(dy)((2x^2-xy+2y^2)/(x-y)) ?

The first d/(dy)((2x^2-xy+2y^2)/(x-y)) is (x^2+4xy-2y^2)/((x-y)^2)

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

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

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

\frac{d}{d y} (x - y) = - 1

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

Simplify

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

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

What is the d/(dy)((2x^2-xy+2y^2)/(x-y)) ?
The d/(dy)((2x^2-xy+2y^2)/(x-y)) is (x^2+4xy-2y^2)/((x-y)^2)
What is the first d/(dy)((2x^2-xy+2y^2)/(x-y)) ?
The first d/(dy)((2x^2-xy+2y^2)/(x-y)) is (x^2+4xy-2y^2)/((x-y)^2)