যদি x=tan ln y হয় তবে y₂/y₁ এর মান কত?

দেওয়া আছে, \(x = \tan(\ln y)\)

এখন, \(x\) এর সাপেক্ষে অন্তরকলন করে পাই,

\(1 = \sec^2(\ln y) \cdot \frac{1}{y} \cdot \frac{dy}{dx}\)

\(\implies \frac{dy}{dx} = y \cos^2(\ln y)\)

\(\implies y_1 = y \cos^2(\ln y)\) ...(1)

আবার, \(y_1\) কে \(x\) এর সাপেক্ষে অন্তরকলন করে পাই,

\(y_2 = \frac{dy_1}{dx} = \frac{d}{dx} [y \cos^2(\ln y)]\)

\(y_2 = \frac{dy}{dx} \cos^2(\ln y) + y \cdot 2\cos(\ln y) \cdot [-\sin(\ln y)] \cdot \frac{1}{y} \cdot \frac{dy}{dx}\)

\(y_2 = y_1 \cos^2(\ln y) - 2\cos(\ln y) \sin(\ln y) y_1/y \)

\(y_2 = y_1 \cos^2(\ln y) - \sin(2\ln y) y_1 \)

\(y_2 = y_1 [\cos^2(\ln y) - \sin(2\ln y) ]\)

এখন, \(\cos^2(\ln y) = \frac{1}{\sec^2(\ln y)} = \frac{1}{1+\tan^2(\ln y)} = \frac{1}{1+x^2}\)

এবং, \(\sin(2\ln y) = \frac{2\tan(\ln y)}{1+\tan^2(\ln y)} = \frac{2x}{1+x^2}\)

অতএব,

\(y_2 = y_1 \left[\frac{1}{1+x^2} - \frac{2x}{1+x^2}\right]\)

\(y_2 = y_1 \left[\frac{1-2x}{1+x^2}\right]\)

সুতরাং,

\(\frac{y_2}{y_1} = \frac{1-2x}{1+x^2} = -\frac{2x-1}{1+x^2}\) ✔️

সুতরাং, \(\frac{y_2}{y_1}\) এর মান \(-\frac{2x-1}{1+x^2}\)।