Derivative Lemma

Question

I am struggling with this and am unsure how to prove it.

Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a differentiable function.

Define $g:\mathbb{R}\rightarrow\mathbb{R}$ by $g(x)=f(x,x,...,x)$ (where $x$ is repeated $n$ times).

How do I prove that:

$$\frac{dg(x)}{dx}=\sum_{i=1}^{n}\bigg[\frac{df(k_1,...,k_n)}{dk_i}\bigg]_{k_i=x}$$

Where $d$ represents the derivative.

Answer 1

Your formula in not quite exact: apply the chain rule in several variables: $$\frac{dg(x)}{dx}=\sum_{i=1}^{n}\frac{\partial f(k_1,...,k_n)}{\partial k_i}\,\frac{d k_i}{dx}$$

Answer 2

I'll do it for $n=2$ (the idea is the same, just more steps adding and subtracting in the general case). You have \begin{align} \frac{g(x+h)-g(x)}h&=\frac{f(x+h,x+h)-f(x,x)}h\\ \ \\ &=\frac{f(x+h,x+h)-f(x,x+h)}h+\frac{f(x,x+h)-f(x,x)}h\\ \ \\ &\to\frac{\partial f}{\partial x_1}f(x,x)+\frac{\partial f}{\partial x_2}f(x,x) \end{align}