If y=f(u) is a differentiable function of u and u=g(x) is a diffe…

Question

If $y=f(u)$ is a differentiable function of $u$ and $u=g(x)$ is a differentiable function of $x$ such that the composite function $y=f[g(x)]$ is a differentlable function of $x$ then prove that
$\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$
Hence find $\frac{d y}{d x}$ if $y=\sqrt{x^2+5}$

✓

Answer

Given: $y=f(u)$ and $u=g(x)$, where $u$ is not a constant function.
Let $\delta x$ the small increment in $x$ and let $\delta u$ and $\delta y$ be the corresponding small increments in $u$ and $y$ respectively.
$\therefore \delta x \neq 0, \delta u \neq 0$ and $\delta y \neq 0$.
We have : $\frac{\delta y}{\delta x}=\frac{\delta y}{\delta u} \cdot \frac{\delta u}{\delta x}$
Taking limit on both sides $\delta x \rightarrow 0$.
$\lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta u}\right) \times \lim _{\delta x \rightarrow u}\left(\frac{\delta u}{\delta x}\right)$
As $\delta x=0$ we get $\delta u \rightarrow 0 \quad$ (as $u$ is a continuous function of $x$ )
$\therefore \lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\lim _{\delta u \rightarrow 0}\left(\frac{\delta y}{\delta u}\right) \times \lim _{\delta x \rightarrow 0}\left(\frac{\delta u}{\delta x}\right)....(1)$
Sine $y$ is a differentiable function of $u$ and $u$ is a differentiable function of $x$,
we get $\lim _{\delta u \rightarrow 0}\left(\frac{\delta y}{\delta u}\right)=\frac{d y}{d u}$ and $\lim _{\overline{\delta x} \rightarrow x}\left(\frac{\delta u}{\delta x}\right)=\frac{d u}{d x}....(2)$
From $(1)$ and $(2)$
$\lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\frac{d y}{d u} \cdot \frac{d u}{d x}....(3)$
The $\text{R.H.S}$. of $(3)$ exists and is finite, it implies that $\text{L.H.S.}$ of $(3)$ also exists and is finite.
$\therefore \lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\frac{d y}{d x}$
$\therefore$ Equation $(i)$ becomes
$\therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}$
Now, $y=\sqrt{x^2+5}$
Let $u=x^2+5$
$\therefore \frac{d u}{d x}=2 x$
Also $y=\sqrt{u}$
$\therefore \frac{d y}{d u}=\frac{1}{2 \sqrt{u}}=\frac{1}{2 \sqrt{x^2+5}}$
$\therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}$
$=\frac{1}{2 \sqrt{x^2+5}}(2 x)$
$=\frac{x}{\sqrt{x^2+5}}$

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