A proof for the chain rule for paths

Let $\gamma: I\rightarrow G$ be a differentiable path and $f:G\rightarrow \mathbb{C}$ a real differentiable function. It is to show, that for the path: $f\circ \gamma: I\rightarrow \mathbb{C}$ the following formula is true:

$\displaystyle{ (f\ \circ \gamma)'(t) = f_{z}(\gamma(t))\gamma'(t)+f_{\overline{z}}(\gamma(t))\overline{\gamma ' (t)}}$

and to conclude that $(f\circ \gamma)'(t)= f'(\gamma(t))\gamma'(t)$.

I began like this:

let $f=u+iv$, it is : $\displaystyle{f_{x}= u_{x}+iv_{x}, f_{y}=u_{y}+iv_{y}, f_{z}=\frac{1}{2}(f_{x}-if_{y}), f_{z}=\frac{1}{2}(f_{x}-if_{y}), f_{\overline{z}}=\frac{1}{2}(f_{x}+if_{y})}$

then:

\begin{align} f \circ \gamma &= f(\gamma(t))= u(x(t),y(t))+iv(x(t),y(t)) \\ (f\ \circ \gamma)'(t) &= u_{x}x'(t)+u_{y}y'(t)+i(v_{x}x'(t)+v_{y}y'(t))\\ &= (u_{x}x'(t)+iv_{x}x'(t))+(u_{y}y'(t)+iv_{y}y'(t))\\ &=f_{x}x'(t) + f_{y}y'(t) \end{align}

stuck .

I don't think this is right so far because I don't see $\overline{\gamma '(t)}$, which should appear. Does anybody see the right way? Please do tell.

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Warning: "Path integral" means something completely different than what you mean. In fact, I don't even know what you mean, since I don't see any integrals at all.

Terminology aside, to prove the identity, just expand: \begin{align} f_z \gamma' + f_{\bar{z}} \bar{\gamma}' &= \frac{1}{2} (f_x - i f_y)(x' + i y') + \frac{1}{2} (f_x + i f_y) (x'-iy') \\ &=\frac{1}{2}( f_x x' + f_y y' + if_x y' - if_y x' + f_x x' + f_y y' - i f_x y' + i f_y x') \\ &= f_x x' + f_y y', \end{align} and you already showed that this last quantity is $(f \circ \gamma)'$.

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Jonathan, Thanks. – VVV Nov 15 '11 at 3:51