4.设 =arcsin (x-y) ,而 x=3t ,=4(t)^3 ,求 dfrac (dz)(dt) .

步骤 1：计算 $\dfrac{\partial z}{\partial x}$ 和 $\dfrac{\partial z}{\partial y}$
由于 $z=\arcsin(x-y)$，根据复合函数的求导法则，我们有：
$\dfrac{\partial z}{\partial x} = \dfrac{1}{\sqrt{1-(x-y)^2}}$
$\dfrac{\partial z}{\partial y} = -\dfrac{1}{\sqrt{1-(x-y)^2}}$
步骤 2：计算 $\dfrac{dx}{dt}$ 和 $\dfrac{dy}{dt}$
由于 $x=3t$ 和 $y=4t^3$，我们有：
$\dfrac{dx}{dt} = 3$
$\dfrac{dy}{dt} = 12t^2$
步骤 3：应用链式法则求 $\dfrac{dz}{dt}$
根据链式法则，我们有：
$\dfrac{dz}{dt} = \dfrac{\partial z}{\partial x} \cdot \dfrac{dx}{dt} + \dfrac{\partial z}{\partial y} \cdot \dfrac{dy}{dt}$
将步骤 1 和步骤 2 的结果代入，我们得到：
$\dfrac{dz}{dt} = \dfrac{1}{\sqrt{1-(x-y)^2}} \cdot 3 + \left(-\dfrac{1}{\sqrt{1-(x-y)^2}}\right) \cdot 12t^2$
$\dfrac{dz}{dt} = \dfrac{3}{\sqrt{1-(x-y)^2}} - \dfrac{12t^2}{\sqrt{1-(x-y)^2}}$
$\dfrac{dz}{dt} = \dfrac{3(1-4t^2)}{\sqrt{1-(3t-4t^3)^2}}$