derivative ln(e^x+xe^x)

解答

导数

ln (e^{x} + x e^{x})

\frac{2 + x}{1 + x}

求解步骤

\frac{d}{d x} (ln (e^{x} + x e^{x}))

使用连锁律:

\frac{1}{e ^{x} + x e ^{x}} \frac{d}{d x} (e^{x} + x e^{x})

= \frac{1}{e ^{x} + x e ^{x}} \frac{d}{d x} (e^{x} + x e^{x})

\frac{d}{d x} (e^{x} + x e^{x}) = 2 e^{x} + e^{x} x

= \frac{1}{e ^{x} + x e ^{x}} (2 e^{x} + e^{x} x)

化简

\frac{1}{e ^{x} + x e ^{x}} (2 e^{x} + e^{x} x) : \frac{2 + x}{1 + x}

= \frac{2 + x}{1 + x}

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

\int_{1}^{6} 8 x d x

d er i v a t i v e \frac{5 x}{1 + x ^{2}}

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

\frac{d}{d x} (- 2 sin (x) + 2 sin (2 x))