# Derivative of $\frac{d}{dt} f(\gamma(t))$ with differential operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \overline{z}}$

Let $$f: \mathbb{C} \rightarrow \mathbb{C}$$ a $$C^1$$ function (i.e. real and imaginary part $$f_1, f_2$$ are continuously differentiable, where $$f=f_1 + i \cdot f_2$$) and let $$\gamma: \mathbb{R} \rightarrow \mathbb{C}, t \mapsto \gamma(t)\, \,$$ $$C^1$$. Then we have that \begin{align} \frac{d}{dt} \, \, f(\gamma(t)) = \frac{\partial f(\gamma(t))}{\partial z} \cdot \gamma'(t) + \frac{\partial f}{\partial \overline{z}} \cdot \overline{\gamma'(t)} \end{align}

I have some trouble to do the calculation to get the formula (I tried to use the Cauchy-Riemann equations but it doesn't work, because the function $$f$$ is not holomorphic). Any suggestion? Thanks in advance!

• I think real and imaginary parts of $f$ will not help you. Perhaps you need $z = \gamma(t)$ and $\overline{z} = \overline{\gamma(t))$. – GEdgar Sep 25 '18 at 19:01

What you are saying is that you have the path in the complex plane that is specified by a parametrization $$\gamma(t)$$. So you have that $$z(t) = \gamma(t) \Rightarrow \bar{z} = \bar{\gamma}$$. Then from the multivariable chain rule you have that
$$\frac{df(z,\bar{z})}{dt} = \frac{\partial f}{\partial z}\frac{d z}{d t} + \frac{\partial f}{\partial \bar{z}}\frac{d \bar{z} }{d t}$$
$$\frac{df(z,\bar{z})}{dt} = \frac{\partial f}{\partial z} \gamma'(t) + \frac{\partial f}{\partial \bar{z}} \bar \gamma'(t)$$
• I think you mean $$\frac{df(z, \bar{z})}{dt} = \frac{\color{red} \partial f}{\color{red} \partial z} \frac{\color{blue} dz}{\color{blue} dt} + \frac{\color{red} \partial f}{\color{red} \partial \bar{z}} \frac{\color{blue} d \bar{z}}{\color{blue} dt}$$ – Mattos Sep 26 '18 at 4:54