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

The answer to (\partial)/(\partial y)(x^y-y^x) is x^yln(x)-xy^{x-1}

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

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

x^{y} ln (x) - x y^{x - 1}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (x^{y}) = x^{y} ln (x)

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

= x^{y} ln (x) - x y^{x - 1}

y = (x^{2} - 7 x + 7) e^{\frac{9 x}{7}}

\int_{0}^{\infty} cos (x) d x

\int (6 x^{2} + 8 x + 3) d x

y^{^{'}} = 8 x^{3} e^{- 2 y}

\frac{d}{d x} (- e^{- \frac{x ^{2}}{2}} x)

What is (\partial)/(\partial y)(x^y-y^x) ?
The answer to (\partial)/(\partial y)(x^y-y^x) is x^yln(x)-xy^{x-1}