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}}

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

y = 4 x - 3 x^{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

\int \frac{8}{4 + x ^{2}} d x

\frac{\partial}{\partial x} (x^{3} e^{x^{2} + y} - 12 y^{2})

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