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

The answer to (\partial)/(\partial x)(x^y+y^x) is yx^{y-1}+y^xln(y)

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{\partial}{\partial x} (2 x^{3} + y)

\int + \frac{4 - 32 x}{1 - 16 x ^{2}} d x

\int \frac{x - 1}{x ( x ^{2} + 1 )} d x

x^{2} \frac{d y}{d x} + x y = 1

\frac{d}{d x} (x^{\frac{2}{3}} (2 x - x^{2}))

What is (\partial)/(\partial x)(x^y+y^x) ?
The answer to (\partial)/(\partial x)(x^y+y^x) is yx^{y-1}+y^xln(y)