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

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

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(\partial}{\partial y}(\frac{xe^x)/y)

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

\frac{1}{s + 3}

y = x^{4} + 9 x^{2} - x (1.9)

2 cos (x) + 3 sin (x), at x = \frac{π}{2}

\frac{ln ( 2 x )}{2 x}

y = 20 x, y = x^{2}

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