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)

4 x^{2}, at x = 3

\frac{\partial}{\partial θ} (θ)

\int 12 x^{2} - sin (x) d x

x^{2} y^{^{'}} - 2 x^{3} e^{\frac{- 1}{x}} = y

\int \frac{4}{x ( x ^{2} + 4 ) ^{2}} d x

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)