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I want to prove that

$\frac{\partial}{\partial x_i}(g \circ f)(x)=g'(f(x))\frac{\partial f}{\partial x_i}(x)$

for a point x in $\mathbb{R}$ assuming that the parial derivatives exist for $f$ and $g \circ f$.

This makes sense to me since its just the normal Chain Rule but with a partial derivative, but how would I prove it?

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1 Answer 1

So you consider a function $f:\mathbb{R}^n\to\mathbb{R}$ and a function $g:\mathbb{R}\to \mathbb{R}$. For the sake of simplicity, assume that they are differentiable everywhere.

Let $\mathbf{x}=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $1\le i\le n$ fixed. What is $\frac{\partial f}{\partial x_i}(x_1,\dots,x_n)$ ? By definition, $$\frac{\partial f}{\partial x_i}(\mathbf{x})=\lim_{x_\to x_i} \frac{f(x_1,\dots,x,\dots,x_n)-f(\mathbf{x})}{x-x_i}.$$ In other words, if you consider the fonction $f_i: \mathbb{R}\to\mathbb{R}$ such that $f_i(x)=f(x_1,\dots,x,\dots,x_n)$, then simply $$\frac{\partial f}{\partial x_i}(\mathbf{x})=f_i'(x_i),$$ the usual derivative.

Now, back to your problem, you are interested in the function $h=g\circ f:\mathbb{R}^n\to\mathbb{R}$ and $\frac{\partial h}{\partial x_i}$. We consider in the same notations as above the function $h_i: \mathbb{R}\to\mathbb{R}$. You can easily see that in fact $h_i=g\circ f_i$. Therefore, by the chain rule for functions from $\mathbb{R}$ to $\mathbb{R}$ (note that $g$ and $f_i$ are two such functions), you get that $$\frac{\partial h}{\partial x_i}(\mathbf{x})=(g\circ f_i)'(x_i)=g'(f_i(x_i))f_i'(x)=g'(f(x))\frac{\partial f}{\partial x_i}(\mathbf{x}).$$

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