Question
If $y=f(u)$ is a differentiable function of $u$ and $u=g(x)$ is a differentlable function of $x$, then prove that $y=f g(x)$ is a differentiable function of $x$ and $\frac{d y}{d x}=\frac{d y}{d u}. \frac{d u}{d x} \quad$
Let δx be a small increment in x.
Let δy and δu be the corresponding increments in y and u respectively
As δx → 0, δy → 0, δu → 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 δx → 0, on both sides,
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\lim _{\delta x \rightarrow 0}\left(\frac{\partial t y}{\delta u} \times \frac{\delta u}{\delta x}\right)$
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta u} \times \lim _{\delta x \rightarrow 0} \frac{\delta u}{\delta x} \ldots(1)$
Since $y$ is a differentiable function of $u , \lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta u}$ exists
and $\lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta x}$ exists as $u$ is a differentiable function of $x$.
Hence, R.H.S. of (1) exists
now $\lim _{\delta u \rightarrow 0} \frac{\delta y}{\delta u}=\frac{d y}{d u}$ and $\lim _{\delta u \rightarrow 0} \frac{\delta u}{\delta x}=\frac{d u}{d x}$
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d u} \times \frac{d u}{d x}$
Since R.H.S. exists, L.H.S. of (1) also exists and
$\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d x}$
$\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$
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