If \(y = f(u)\) is a differentiable function of \(u\) and \(u = g(x)\) is a differentiable function of \(x\) then prove that \(y\) is a differentiable function of \(x\) and \(\dfrac{dy}{dx} = \dfrac{dy}{du}\times\dfrac{du}{dx}\).

Answer: Proof of the chain rule.

Let \(\delta x\) be a small increment in \(x\), and let \(\delta u\) and \(\delta y\) be the corresponding increments in \(u\) and \(y\). We assume \(\delta u \ne 0\) for small \(\delta x \ne 0\).

Since \(u = g(x)\) is differentiable, it is continuous, so \(\delta u \to 0\) as \(\delta x \to 0\).

Now \(\dfrac{\delta y}{\delta x} = \dfrac{\delta y}{\delta u}\times\dfrac{\delta u}{\delta x}\).

Taking limits as \(\delta x \to 0\) (so that \(\delta u \to 0\)):

\(\displaystyle\lim_{\delta x \to 0}\frac{\delta y}{\delta x} = \lim_{\delta u \to 0}\frac{\delta y}{\delta u}\times\lim_{\delta x \to 0}\frac{\delta u}{\delta x}\) ... (1)

Since \(y = f(u)\) is differentiable, \(\displaystyle\lim_{\delta u \to 0}\frac{\delta y}{\delta u} = \frac{dy}{du}\) exists, and since \(u = g(x)\) is differentiable, \(\displaystyle\lim_{\delta x \to 0}\frac{\delta u}{\delta x} = \frac{du}{dx}\) exists.

So the right side of (1) exists and is finite; hence \(\displaystyle\lim_{\delta x \to 0}\frac{\delta y}{\delta x}\) exists, i.e. \(y\) is a differentiable function of \(x\), and \(\dfrac{dy}{dx} = \dfrac{dy}{du}\times\dfrac{du}{dx}\).