d/dx((1-cosx)/(1+cosx))=?

আমরা \(\frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right)\) এর মান বের করতে চাই।

ধরি, \(u = 1 - \cos x\) এবং \(v = 1 + \cos x\)।

তাহলে, \(\frac{du}{dx} = \sin x\) এবং \(\frac{dv}{dx} = -\sin x\)।

ভাগফল বিধি (Quotient Rule) অনুসারে,

\(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

সুতরাং,

\(\frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right) = \frac{(1+\cos x)(\sin x) - (1-\cos x)(-\sin x)}{(1+\cos x)^2}\)

\(= \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{(1+\cos x)^2}\)

\(= \frac{2\sin x}{(1+\cos x)^2}\)

অতএব, \(\frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right) = \frac{2\sin x}{(1+\cos x)^2}\) 🥳