প্রশ্ন: d/dx ln(sinx/(1-cosx)) = ?

আমরা \(ln(\frac{sinx}{1-cosx})\) এর অন্তরকলন নির্ণয় করব।

ধরি, \(y = ln(\frac{sinx}{1-cosx})\)

প্রথমত, লগারিদমের নিয়ম ব্যবহার করে সরল করি:

\(y = ln(sinx) - ln(1-cosx)\)

এখন, x এর সাপেক্ষে অন্তরকলন করি:

\(\frac{dy}{dx} = \frac{d}{dx} [ln(sinx) - ln(1-cosx)]\)

\(\frac{dy}{dx} = \frac{d}{dx} ln(sinx) - \frac{d}{dx} ln(1-cosx)\)

আমরা জানি, \(\frac{d}{dx} ln(f(x)) = \frac{f'(x)}{f(x)}\)

সুতরাং,

\(\frac{dy}{dx} = \frac{cosx}{sinx} - \frac{-sinx}{1-cosx}\)

\(\frac{dy}{dx} = cotx + \frac{sinx}{1-cosx}\)

এখন, যোগ করি:

\(\frac{dy}{dx} = \frac{cosx(1-cosx) + sinx \cdot sinx}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx(1-cosx)}\)

আমরা জানি, \(sin^2x + cos^2x = 1\), সুতরাং \(sin^2x = 1 - cos^2x\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + 1 - cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + sin^2x - cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - (cos^2x - sin^2x)}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos2x}{sinx(1-cosx)}\)

আমরা ত্রিকোণমিতিক সূত্র থেকে পাই, \(1 - cos^2x=sin^2x\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx - sinxcosx}\)

\(\frac{dy}{dx} = \frac{cosx + (1-cos^2x) - cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + 1 - 2cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + sin^2x - cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + 1 - cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + 1 - 2cos^2x}{sinx(1-cosx)}\)

এখন অন্যভাবে,

\(\frac{dy}{dx} = cotx + \frac{sinx}{1-cosx}\)

\(\frac{dy}{dx} = \frac{cosx}{sinx} + \frac{sinx}{1-cosx}\)

\(\frac{dy}{dx} = \frac{cosx(1-cosx) + sin^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + (sin^2x - cos^2x)}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + (1 - cos^2x - cos^2x)}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx +1 - 2cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + (sin^2x-cos^2x)}{sinx(1-cosx)} = \frac{cosx + 1 - 2cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx(1-cosx) + sin^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx - sinxcosx}\)

\(\frac{dy}{dx} = \frac{cosx + 1-2cos^2x}{sinx - sinxcosx}\)

\(\frac{dy}{dx} = \frac{cosx + 1 - 2cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos^2x + sin^2x}{sinx(1-cosx)} = \frac{cosx - (cos^2x - sin^2x)}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx - cos2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + sin^2x - cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + (1 - cos^2x) - cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + 1 - 2cos^2x}{sinx(1-cosx)}\)

\(\frac{dy}{dx} = \frac{cosx + 1 - 2cos^2x}{sinx(1-cosx)}\)

\(\frac{sinx}{1-cosx} = \frac{2sin(x/2)cos(x/2)}{2sin^2(x/2)} = cot(x/2)\)

\(ln(cot(x/2)) = ln(cos(x/2)) - ln(sin(x/2))\)

\(\frac{d}{dx}ln(cot(x/2)) = \frac{-sin(x/2)}{2cos(x/2)} - \frac{cos(x/2)}{2sin(x/2)} = -\frac{1}{2}(tan(x/2)+cot(x/2))\)

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

\(= -\frac{1}{sin(x)} = -cosecx\)

অতএব, \(\frac{dy}{dx} = -cosecx\). 🥳