6.设u=(e^ax(y-z))/(a^2)+1，而y=asinx,z=cosx,求(du)/(dx).

首先，计算 $u$ 对 $x$、$y$、$z$ 的偏导数： \[ \frac{\partial u}{\partial x} = \frac{ae^{ax}(y-z)}{a^2+1}, \quad \frac{\partial u}{\partial y} = \frac{e^{ax}}{a^2+1}, \quad \frac{\partial u}{\partial z} = -\frac{e^{ax}}{a^2+1}. \] 然后，计算 $y$ 和 $z$ 对 $x$ 的导数： \[ \frac{dy}{dx} = a \cos x, \quad \frac{dz}{dx} = -\sin x. \] 应用链式法则： \[ \frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \cdot \frac{dy}{dx} + \frac{\partial u}{\partial z} \cdot \frac{dz}{dx}. \] 代入并化简： \[ \frac{du}{dx} = \frac{e^{ax}}{a^2+1} \left[ a(y-z) + a \cos x + \sin x \right]. \] 将 $y = a \sin x$，$z = \cos x$ 代入： \[ \frac{du}{dx} = \frac{e^{ax}}{a^2+1} \left[ a^2 \sin x - a \cos x + a \cos x + \sin x \right] = e^{ax} \sin x. \] **答案：** \[ \boxed{e^{ax} \sin x} \]