What is (\partial)/(\partial x)((xz+yz)/(xy)) ?

The answer to (\partial)/(\partial x)((xz+yz)/(xy)) is -z/(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)((xz+yz)/(xy))

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

- \frac{z}{x ^{2}}

Solution steps

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

Treat

z, y

as constants

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{\partial}{\partial x} (x z + yz) = z

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

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

Simplify

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

= - \frac{z}{x ^{2}}

f (x) = \frac{x}{2 + x}

\frac{d}{d x} (\frac{A}{x})

x \to 1 lim (x^{2} + 2 x - \frac{5}{2})

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0, y (0) = 8

\int_{- 1}^{1} e^{y} - (y^{2} - 7) d y

What is (\partial)/(\partial x)((xz+yz)/(xy)) ?
The answer to (\partial)/(\partial x)((xz+yz)/(xy)) is -z/(x^2)