What is (\partial)/(\partial x)((25xy)/(x-y)) ?

The answer to (\partial)/(\partial x)((25xy)/(x-y)) is -(25y^2)/((x-y)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)((25xy)/(x-y))

\frac{\partial}{\partial x} (\frac{25 x y}{x - y})

- \frac{25 y ^{2}}{( x - y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{25 x y}{x - y})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 25 y \frac{\partial}{\partial x} (\frac{x}{x - y})

Apply the Quotient Rule:

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

= 25 y \frac{\frac{\partial}{\partial x} ( x ) ( x - y ) - \frac{\partial}{\partial x} ( x - y ) x}{( x - y ) ^{2}}

\frac{\partial}{\partial x} (x) = 1

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

= 25 y \frac{1 \cdot ( x - y ) - 1 \cdot x}{( x - y ) ^{2}}

Simplify

25 y \frac{1 \cdot ( x - y ) - 1 \cdot x}{( x - y ) ^{2}} : - \frac{25 y ^{2}}{( x - y ) ^{2}}

= - \frac{25 y ^{2}}{( x - y ) ^{2}}

tan (\frac{π x}{2})

y^{^{''}} + 4 y = - 5 csc (2 t)

\int \frac{1}{3 - x ^{2}} d x

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

\int_{0}^{2} 0.5 x d x

What is (\partial)/(\partial x)((25xy)/(x-y)) ?
The answer to (\partial)/(\partial x)((25xy)/(x-y)) is -(25y^2)/((x-y)^2)