$y=sin^{-1}\frac{4x}{1+4x^{2}}$ হলে $\frac{dx}{dx}$ এব মান

এখানে $y=\sin^{-1}\frac{4x}{1+4x^{2}}$ দেওয়া আছে। আমাদের $\frac{dy}{dx}$ নির্ণয় করতে হবে। আমরা জানি, $\frac{d}{dx}(\sin^{-1}u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$ । এখানে, $u = \frac{4x}{1+4x^{2}}$ । প্রথমে $\frac{du}{dx}$ নির্ণয় করি। ভাগফলের সূত্র ব্যবহার করে পাই: $\frac{du}{dx} = \frac{(1+4x^{2})\frac{d}{dx}(4x) - (4x)\frac{d}{dx}(1+4x^{2})}{(1+4x^{2})^{2}}$ $\frac{du}{dx} = \frac{(1+4x^{2})(4) - (4x)(8x)}{(1+4x^{2})^{2}}$ $\frac{du}{dx} = \frac{4+16x^{2} - 32x^{2}}{(1+4x^{2})^{2}}$ $\frac{du}{dx} = \frac{4-16x^{2}}{(1+4x^{2})^{2}}$ এবার, $\frac{dy}{dx}$ নির্ণয় করি: $\frac{dy}{dx} = \frac{1}{\sqrt{1-(\frac{4x}{1+4x^{2}})^{2}}} \cdot \frac{4-16x^{2}}{(1+4x^{2})^{2}}$ $\frac{dy}{dx} = \frac{1}{\sqrt{\frac{(1+4x^{2})^{2}-(4x)^{2}}{(1+4x^{2})^{2}}}} \cdot \frac{4-16x^{2}}{(1+4x^{2})^{2}}$ $\frac{dy}{dx} = \frac{1+4x^{2}}{\sqrt{1+8x^{2}+16x^{4}-16x^{2}}} \cdot \frac{4-16x^{2}}{(1+4x^{2})^{2}}$ $\frac{dy}{dx} = \frac{1+4x^{2}}{\sqrt{1-8x^{2}+16x^{4}}} \cdot \frac{4-16x^{2}}{(1+4x^{2})^{2}}$ $\frac{dy}{dx} = \frac{1+4x^{2}}{\sqrt{(1-4x^{2})^{2}}} \cdot \frac{4-16x^{2}}{(1+4x^{2})^{2}}$ $\frac{dy}{dx} = \frac{1+4x^{2}}{|1-4x^{2}|} \cdot \frac{4(1-4x^{2})}{(1+4x^{2})^{2}}$ যদি $1-4x^{2}>0$ হয়, তবে $|1-4x^{2}| = 1-4x^{2}$ : $\frac{dy}{dx} = \frac{1+4x^{2}}{1-4x^{2}} \cdot \frac{4(1-4x^{2})}{(1+4x^{2})^{2}}$ $\frac{dy}{dx} = \frac{4}{1+4x^{2}}$ এই ফলাফল বিকল্প 2-এর সাথে মিলে যায়।