The partial derivative of a composed function.

Question

I think I have either found an error in my notes either I misunderstood something.

Suppose:

$$f:\mathbb{R}^n \rightarrow \mathbb{R}^m $$ differentiable at $x_0$ $$g:\mathbb{R}^m \rightarrow \mathbb{R}^p $$ differentiable at $f(x_0)$

Now suppose $h: \mathbb{R}^n \rightarrow \mathbb{R}^p$ such that: $$h: x \mapsto g \circ f(x) $$

Then $h$ is differentiable at $x_0$ and $$\frac{\partial h_i}{\partial x_j}(x_0)= \sum_{k=1}^{m}\frac{\partial g}{\partial x_j}(f(x_0))\frac{\partial f_k}{\partial x_j}(x_0) $$

My question is why do we have $\sum_{k=1}^{m}\frac{\partial g}{\partial x_j}(f(x_0))$ and not $$\sum_{k=1}^{m}\frac{\partial g_k}{\partial x_j}(f(x_0))$$

And overall, is this formula even correct?

Answer 1

No, the correct formula is $$ \frac{\partial h_i}{\partial x_j}(x_0) = \sum_{k=1}^{m}\frac{\partial g_{\color{red}i}}{\partial x_{\color{red}k}}(f(x_0))\frac{\partial f_k}{\partial x_j}(x_0) , $$ or maybe better (to avoid confusion) $$ \frac{\partial h_i}{\partial x_j}(x_0) = \sum_{k=1}^{m}\frac{\partial g_{\color{red}i}}{\partial \color{red}{y_k}}(f(x_0))\frac{\partial f_k}{\partial x_j}(x_0) , $$ if we write $g=g(y_1,\dots,y_m)$ instead of $g=g(x_1,\dots,x_m)$.