What is (\partial}{\partial y}(\frac{2x)/y) ?

The answer to (\partial}{\partial y}(\frac{2x)/y) is -(2x)/(y^2)

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Simplify

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

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

\frac{d y}{d x} - \frac{( 2 y )}{( 3 x )} + \frac{( x ^{2} + 2 )}{3} \cdot y^{4} = 0

\int e^{x} - ln (x) d x

\int \frac{x ^{2}}{4 + x ^{6}} d x

x^{^{''}} + 3 x = 0

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

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