If \(y=f(u)\) is a differentiable function of \(u\) and \(u=g(x)\) is differentiable function of \(x\) such that the composite function \(y=f[g(x)]\) is a differentiable function of \(x\) then prove that : \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{…

Answer: Proved; \(\dfrac{d}{dx}\left(\dfrac1{\sqrt{\sin x}}\right)=-\dfrac{\cos x}{2(\sin x)^{3/2}}\)

Proof: Let \(\delta x\) be a small increment in \(x\), and \(\delta u\), \(\delta y\) the corresponding increments in \(u\) and \(y\). Since \(u=g(x)\) is differentiable, it is continuous, so \(\delta u\to0\) as \(\delta x\to0\). Assume \(\delta u\ne0\) for small \(\delta x\).

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

Taking limits as \(\delta x\to0\) (so \(\delta u\to0\)):

\(\displaystyle\lim_{\delta x\to0}\dfrac{\delta y}{\delta x}=\lim_{\delta u\to0}\dfrac{\delta y}{\delta u}\times\lim_{\delta x\to0}\dfrac{\delta u}{\delta x}\).

Since \(y\) is a differentiable function of \(u\) and \(u\) of \(x\), both limits on the right exist and equal \(\dfrac{dy}{du}\) and \(\dfrac{du}{dx}\). Hence the left limit exists and \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\).

Hence part: Let \(y=\dfrac1{\sqrt{\sin x}}=u^{-1/2}\) where \(u=\sin x\).

\(\dfrac{dy}{du}=-\dfrac12u^{-3/2}\), \(\dfrac{du}{dx}=\cos x\).

\(\dfrac{dy}{dx}=-\dfrac12(\sin x)^{-3/2}\cos x=-\dfrac{\cos x}{2(\sin x)^{3/2}}\) \(\left(=-\dfrac12\cot x\sqrt{\operatorname{cosec}x}\right)\).