What is (\partial)/(\partial y)((x+y)/(xy)) ?

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

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

- \frac{1}{y ^{2}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= - \frac{1}{y ^{2}}

y^{^{''}} - 5 y^{^{'}} - 6 y = 0

\int 51 - 14 x - x^{2} d x

\int \frac{x ^{3} - 8}{x ^{11}} d x

\frac{\partial}{\partial x} (ln (x^{8} - y^{6}))

\int (x^{2} - 9)^{4} d x

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