What is (\partial)/(\partial x)(1/(xy^2-x^2y)) ?

The answer to (\partial)/(\partial x)(1/(xy^2-x^2y)) is -(y-2x)/(x^2y(y-x)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(1/(xy^2-x^2y))

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 lim ((e^{x} + x)^{\frac{1}{x}})

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

x \to \infty lim (\frac{( ln ( e ^{3 x} + x ) )}{x})

t \frac{d y}{d t} + 2 y = t^{4}

\frac{\partial}{\partial y} (x ln (y) - y ln (x))

What is (\partial)/(\partial x)(1/(xy^2-x^2y)) ?
The answer to (\partial)/(\partial x)(1/(xy^2-x^2y)) is -(y-2x)/(x^2y(y-x)^2)