What is (\partial)/(\partial x)((2x-3)/(xy)) ?

The answer to (\partial)/(\partial x)((2x-3)/(xy)) is 3/(x^2y)

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(\partial)/(\partial x)((2x-3)/(xy))

\frac{\partial}{\partial x} (\frac{2 x - 3}{x y})

\frac{3}{x ^{2} y}

Solution steps

\frac{\partial}{\partial x} (\frac{2 x - 3}{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 - 3}{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 - 3 ) x - \frac{\partial}{\partial x} ( x ) ( 2 x - 3 )}{x ^{2}}

\frac{\partial}{\partial x} (2 x - 3) = 2

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

= \frac{1}{y} \cdot \frac{2 x - 1 \cdot ( 2 x - 3 )}{x ^{2}}

Simplify

\frac{1}{y} \cdot \frac{2 x - 1 \cdot ( 2 x - 3 )}{x ^{2}} : \frac{3}{x ^{2} y}

= \frac{3}{x ^{2} y}

7 - 3 u

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

y^{^{''}} + y^{^{'}} - 2 y = 2 t, y (0) = 0, y^{^{'}} (0) = 7

y = x^{3} - 2 x + 2 (2.6)

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

What is (\partial)/(\partial x)((2x-3)/(xy)) ?
The answer to (\partial)/(\partial x)((2x-3)/(xy)) is 3/(x^2y)