Question

If $y=f\left(u\right)$ is a differentiable function of $u$ and $u=g\left(x\right)$ is a differentiable function of $x$, then prove that $y=f\left(g\right(x\left)\right)$ is a differentiable function of $x$ and $\frac{dy}{dx}=\frac{dy}{du}\times \frac{dy}{dx}$

Answer 1

Let $\delta x,\delta y$ and $\delta u$ be a small change in $x,y$ and $u$ respectively.
As, $\delta x\to 0,\delta y\to 0,\delta u\to 0$.
As $u$ is differentiable function, it is continuous.
Consider the incrementary ratio $\frac{\delta y}{\delta x}$
We have, $\frac{\delta y}{\delta x}=\frac{\delta y}{\delta u}\times \frac{\delta u}{\delta x}$
Taking limit as $\delta x\to 0$ on both sides
$\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}=\underset{\delta x\to 0}{lim}(\frac{\delta y}{\delta u}\times \frac{\delta u}{\delta x})$
$\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}=\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta u}\times \underset{\delta x\to 0}{lim}\frac{\delta u}{\delta x}$
$\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}=\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta y}\times \underset{\delta x\to 0}{lim}\frac{\delta u}{\delta x}.....\left(i\right)$
Since $y$ is a differentiable function of $u,\underset{\delta x\to 0}{lim}\left(\frac{\delta y}{\delta u}\right)$ exists and $\underset{\delta x\to 0}{lim}\left(\frac{\delta u}{\delta x}\right)$ exists as $u$ is a differentiable function of $x$.
R.H.S. of equation (i) exists.
Now, $\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta u}=\frac{dy}{du}$ and $\underset{\delta x\to 0}{lim}\frac{\delta u}{\delta x}=\frac{du}{dx}$
$\therefore \underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}=\frac{dy}{du}\times \frac{du}{dx}$
Since R.H.S. exists.
$\therefore$ L.H.S. also exists and $\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}=\frac{dy}{dx}$
$\therefore \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$