Why $g^{\prime}(\gamma(t)) = \left(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\right)$ in Complex Analysis by Cain (ch4, pg 4.7)

Question

At page 4.7 in chapter 4 of this link, the author writes:

$$\frac{d}{dt} g(\gamma(t)) = \left(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\right) \left(\frac{\partial x}{\partial t} + i \frac{dy}{dt}\right) = g^{\prime}(\gamma(t))\gamma^{\prime}(t)$$

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I understand that $\gamma^{\prime}(t) = \left(\frac{\partial x}{\partial t} + i \frac{dy}{dt}\right)$ because $\gamma(t)$ is defined as $x(t) + i y(t)$.

What I don't follow is why $g^{\prime}(\gamma(t)) = \left(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\right)$.

$$g(\gamma(t)) = u(x(t),y(t)) + i v(x(t),y(t))$$ so why are the partials only taken with respect to $x$ and not also $y$? Thanks

Answer 1

$g:U \to \mathbb{C}$ where $U $ is a open subset of $\mathbb{C}$.Let $g=u+iv $ and if the partial derivatives $u_x,u_y,v_x,v_y$ exists and are continuous then $g$ is differentiable and $g'= u_x^{'}+iv_x^{'}$. Similarly $g'= u_y^{'}+iv_y^{'}$.

The author could have used this result and wrote like this...