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

解答

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

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

求解步骤

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

将

x

视为常数

使用微分加减法定则:

(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}

d er i v a t i v e 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)