y=tan^-1((4x)/(1-4x^2))  হলে dy/dx= কত ?

Another Explanation (5): প্রশ্ন: \( y = \tan^{-1} \left( \frac{4x}{1 - 4x^2} \right) \) হলে \(\frac{dy}{dx}\) কত? সমাধান: প্রথমে, \( y \) এর জন্য ডিফারেনশিয়েশন করতে হবে: \[ y = \tan^{-1} \left( u \right), \quad \text{যেখানে} \quad u = \frac{4x}{1 - 4x^2} \] \[ \Rightarrow \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] এখন, \( u = \frac{4x}{1 - 4x^2} \) এর জন্য ডিফারেনশিয়েশন: \[ \frac{du}{dx} = \frac{(4)(1 - 4x^2) - 4x \cdot (-8x)}{(1 - 4x^2)^2} \] \[ = \frac{4(1 - 4x^2) + 32x^2}{(1 - 4x^2)^2} \] \[ = \frac{4 - 16x^2 + 32x^2}{(1 - 4x^2)^2} \] \[ = \frac{4 + 16x^2}{(1 - 4x^2)^2} \] এখন, \( u^2 \) এর জন্য: \[ u^2 = \left( \frac{4x}{1 - 4x^2} \right)^2 = \frac{16x^2}{(1 - 4x^2)^2} \] অতএব, \[ 1 + u^2 = 1 + \frac{16x^2}{(1 - 4x^2)^2} = \frac{(1 - 4x^2)^2 + 16x^2}{(1 - 4x^2)^2} \] \[ = \frac{(1 - 4x^2)^2 + 16x^2}{(1 - 4x^2)^2} \] বর্গের বিস্তার করি: \[ (1 - 4x^2)^2 = 1 - 8x^2 + 16x^4 \] অতএব, \[ 1 + u^2 = \frac{1 - 8x^2 + 16x^4 + 16x^2}{(1 - 4x^2)^2} = \frac{1 + (-8x^2 + 16x^2) + 16x^4}{(1 - 4x^2)^2} \] \[ = \frac{1 + 8x^2 + 16x^4}{(1 - 4x^2)^2} \] এখন, \[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} = \frac{(1 - 4x^2)^2}{1 + 8x^2 + 16x^4} \times \frac{4 + 16x^2}{(1 - 4x^2)^2} \] নোট করি যে, \((1 - 4x^2)^2\) ক্যান্সেল হয়ে যাবে: \[ \frac{dy}{dx} = \frac{4 + 16x^2}{1 + 8x^2 + 16x^4} \] নোট করি, \[ 4 + 16x^2 = 4(1 + 4x^2) \] এবং, \[ 1 + 8x^2 + 16x^4 = (1 + 4x^2)^2 \] অতএব, \[ \frac{dy}{dx} = \frac{4(1 + 4x^2)}{(1 + 4x^2)^2} = \frac{4}{1 + 4x^2} \] ??ত্তর: \[ \boxed{\frac{dy}{dx} = \frac{4}{1 + 4x^2}} \]