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}

\frac{\partial}{\partial x} (1 + cos (y))

x \to 0 lim (1 - cos^{3} (x))

(2 y - x^{3}) d x = x d y

6 y^{^{'}} + 5 y = 5

x \to 2 lim (x^{2})

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}