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

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

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

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

\frac{1}{x ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

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

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

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

Simplify

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

= \frac{1}{x ^{2}}

\frac{d}{d x} (x^{2} + 4 x sin (x))

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

\int t t - 1 d t

\int \frac{1}{x 4 x ^{2} + 1} d x

(cosh (x))^{^{'}}

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