3.3 Chain rule

$f \circ g = f \circ g(x) = f(g(x)) = f(g)$

$\frac{d(f \circ g)}{dx} = \frac{d(f(g))}{dx} = f^{\prime} \cdot g^{\prime} = f^{\prime}(g) \cdot g^{\prime}(x) = \frac{df}{dg} \frac{dg}{dx}$

Proof. $u = g(x), \quad k = g(x + h) - g(x) = g(x + h) - u, \quad u + k = g(x + h)$

$f^{\prime}(u) = \lim\limits_{k \to 0} \frac{f(u + k) - f(u)}{k}$

$\varphi(k) = \frac{f(u + k)- f(u)}{k} - f^{\prime}, \quad \lim\limits_{k \to 0} \varphi(k) = 0$

$k \varphi(k) = f(u + k) - f(u) - kf^{\prime}(u)$

$f(u + k) - f(u) = k \varphi(k) + kf^{\prime}(u)$

$\frac{f(g(x + h)) - f(g(x))}{h} = \frac{f(u + k) - f(u)}{h} = \frac{kf^{\prime}(u) + k\varphi(k)}{h}$

$= \frac{k}{h}f^{\prime}(u) + \frac{k}{h}\varphi(k) = \frac{g(x + h) - g(x)}{h}f^{\prime}(u) + \frac{g(x + h) - g(x)}{h}\varphi(k)$

$\frac{d(f(g))}{dx} = \lim\limits_{h \to 0} \frac{f(g(x + h)) - f(g(x))}{h} = \lim\limits_{h \to 0} \left[ \frac{g(x + h) - g(x)}{h} f^{\prime} + \frac{g(x + h) - g(x)}{h}\varphi(k) \right]$

$= g^{\prime}(x) \cdot f^{\prime}(u) + g^{\prime}(x) \cdot 0$

$= f^{\prime}(u) \cdot g^{\prime}(x)$

$\frac{d}{dx}(f \circ g \circ h) = \frac{d(f \circ g \circ h)}{dx} = \frac{df}{dg} \frac{dg}{dh} \frac{dh}{dx}$