What is (\partial)/(\partial y)(x^2e^{-x/y}) ?

The answer to (\partial)/(\partial y)(x^2e^{-x/y}) is (e^{-x/y}x^3)/(y^2)

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

x^{2} e^{- \frac{x}{y}} \frac{x}{y ^{2}} : \frac{e ^{- \frac{x}{y}} x ^{3}}{y ^{2}}

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

\int 13 e^{u} d u

f (x) = x^{2} - 3 x + 5, at x = 2

y^{^{'}} = \frac{y}{x} + 1 + (\frac{y}{x})^{2}

\int \frac{x ^{2} - 12}{x ^{3} + 4 x} d x

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