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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{2}{( s - 2 ) ^{2}}

y^{3} - (10 x + 8) + 3 x y^{2} y^{^{'}} = 0

\frac{x - 4}{x + 4}

y = \frac{1}{3 x}

\frac{d}{d x} (sin (x^{3} - x))

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