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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{d}{d x} (a ln (x))

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\frac{- 3 x ^{4} + 1}{( x ^{4} + 1 ) ^{2}}

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