这是一份liverpool利物浦大学MATH243的成功案例

$$
\frac{d w}{d z}=\frac{\delta w}{8 x}=\frac{1}{i} \frac{\delta v}{\delta y}
$$
follow
und
$$
\frac{\delta}{\delta c}\left(\frac{d w}{d z}\right)=\frac{1}{i} \frac{\delta^{2} w}{\delta x \delta y}
$$
$$
\frac{\delta}{\delta y}\left(\frac{d v}{d z}\right)=\frac{\partial^{2} v}{8 x \delta y}
$$
sonsequently
$$
\frac{\delta}{\delta y}\left(\frac{d u t}{d z}\right)=i \frac{\delta}{\delta x}\left(\frac{d u t}{d z}\right) \text {, }
$$

MATH243 COURSE NOTES ：

$$
w^{3}-w+z=0
$$
If, for brevity, we put
$$
p=\sqrt[3]{\frac{1}{2}\left(-z-\sqrt{z^{2}-\frac{4}{27}}\right)}, q=\sqrt[3]{\frac{1}{2}\left(-z+\sqrt{\left.z^{2}-\frac{4}{27}\right)}\right.},
$$
and the two imaginary cube roots of unity
$$
\frac{-1+i \sqrt{3}}{2}=\alpha, \frac{-1-i \sqrt{3}}{2}=\alpha^{2}
$$
Cardan’s formula gives for the three roots of the above