What is (\partial)/(\partial x)((4-x^2y)/(2x)) ?

The answer to (\partial)/(\partial x)((4-x^2y)/(2x)) is (-x^2y-4)/(2x^2)

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(\partial)/(\partial x)((4-x^2y)/(2x))

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

\frac{- x ^{2} y - 4}{2 x ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= \frac{1}{2} \frac{\partial}{\partial x} (\frac{4 - x ^{2} y}{x})

Apply the Quotient Rule:

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

= \frac{1}{2} \cdot \frac{\frac{\partial}{\partial x} ( 4 - x ^{2} y ) x - \frac{\partial}{\partial x} ( x ) ( 4 - x ^{2} y )}{x ^{2}}

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

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

= \frac{1}{2} \cdot \frac{( - 2 y x ) x - 1 \cdot ( 4 - x ^{2} y )}{x ^{2}}

Simplify

\frac{1}{2} \cdot \frac{( - 2 y x ) x - 1 \cdot ( 4 - x ^{2} y )}{x ^{2}} : \frac{- x ^{2} y - 4}{2 x ^{2}}

= \frac{- x ^{2} y - 4}{2 x ^{2}}

\frac{\partial}{\partial x} (3 x y^{2} + 20 x^{2} y^{3})

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{4}) = 2, y^{^{'}} (\frac{π}{4}) = - 2

\int \frac{3 x ^{2} + 17 x + 32}{x ^{3} + 8 x ^{2} + 16 x} d x

y^{^{''}} = 1 + (y^{^{'}})^{2}

What is (\partial)/(\partial x)((4-x^2y)/(2x)) ?
The answer to (\partial)/(\partial x)((4-x^2y)/(2x)) is (-x^2y-4)/(2x^2)