Proving the composition of two functions having partial derivatives has a partial derivative.

Let $N$ be open subset of $\Bbb R^n$, $x \in N$

The function $f : N \to \Bbb R$ has a partial derivative at point $x$

Let $I$ be open interval in $\Bbb R$ with $f(N) \subset I $

The function $g: I \to \Bbb R$ have a derivative at $f(x)$

How do I prove that the compositon $g\circ f : N \to \Bbb R $ has a partial derivative at $x$

$\frac {\partial}{\partial x_i}(g\circ f)(x) = g'(f(x))\frac{\partial f}{\partial x_i}(x)$ with component $i$?

Please show me a clear way to solve this.

Solution 1:

Consider $h = g \circ f$. Let us calculate the partial derivative at $p \in N $ with respect to the $j$-th coordinate. Observe that $\gamma(t) = p+te_j$ gives a line in the $j$-th coordinate direction for which $\gamma(0)=p$. The partial derivative of $h$ with respect to $x_j$ at $p$ is defined by $$ \frac{\partial h}{\partial x_j}(p)=\frac{d}{dt} \bigl( h \circ \gamma \bigr)(t) |_{t=0} =\lim_{t \rightarrow 0} \frac{ h(p+te_j)-h(p)}{t} $$ Consider, $$ \frac{d}{dt} \bigl( h \circ \gamma \bigr)(t) = \frac{d}{dt} g( f (\gamma(t))) = g'(f(\gamma(t)) \frac{d}{dt}f(\gamma(t) ) $$ where I used the single-variate chain rule in the last equality. Now, if you understand the definition I used for the partial derivative, your result is obvious from what I've written. Although, I replaced $x$ with $p$.