সৃজনশীলঃ ‘দৃশ্যকল্প-১ u(x)=cos⁡(z)এবং v(x)=xsin−1x…’ নটরডেম কলেজ ২০২৪ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $v\left(x\right)=x^{{\sin }^{-1}x}$বা, $\ln v={\sin }^{-1}x\cdot \ln x$বা, $\frac{1}{v}\frac{dv}{dx}=\frac{{\sin }^{-1}x}{x}+\frac{\ln x}{\sqrt{1-x^2}}$ [অন্তরীকরণ করে]

$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=x^{{\sin }^{-1}x}(\frac{{\sin }^{-1}x}{x}+\frac{\ln x}{\sqrt{1-x^2}})$

আবার,

$u\left(x\right)=\cos z=\cos ({\cot }^{-1}\sqrt{r\left(x\right)})=\cos ({\cot }^{-1}\sqrt{\frac{1-x}{1+x}})$

ধরি, $m\left(x\right)=\left\{u\right(x){\}}^4={\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\}$

এবং $x=\cos 2\theta$

$\therefore m\left(x\right)={\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\}$

$={\cos }^4\{{\cot }^{-1}\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}\}={\cos }^4\{{\cot }^{-1}\cdot \tan \theta \}$

$={\cos }^4\{{\cot }^{-1}\cot (\frac{\pi }{2}-\theta )\}={\cos }^4(\frac{\pi }{2}-\theta )$

$={\sin }^4\theta =\frac{1}{4}{(2{\sin }^2\theta )}^2=\frac{1}{4}(1-\cos 2\theta {)}^2$

$=\frac{1}{4}(1-x{)}^2[\because x=\cos 2\theta ]$

$\therefore m\left(x\right)={\cos }^4\{{\cot }^{-1}\sqrt{\frac{1-g\left(x\right)}{1+g\left(x\right)}}\}=\frac{1}{4}(1-x{)}^2$

$\frac{\mathrm{dm}}{\mathrm{dx}}=\frac{1}{4}\cdot 2(1-x)(-1)=\frac{1}{2}(x-1)$

প্রদত্ত রাশি, $y=\left\{u\right(x){\}}^4+v(x)$

বা, $y=m\left(x\right)+v\left(x\right)$ বা, $\frac{dy}{dx}=\frac{dm}{dx}+\frac{dv}{dx}$

$\therefore \frac{dy}{dx}=\frac{1}{2}(x-1)+x^{{\sin }^{-1}x}(\frac{{\sin }^{-1}x}{x}+\frac{\ln x}{\sqrt{1-x^2}})$ (Ans.)$

দেওয়া আছে, $f\left(x\right)={\tan }^{-1}x$

$\therefore y=\tan \left\{\mathrm{mf}\right(x\left)\right\}=\tan \left({\mathrm{mtan}}^{-1}x\right)$

বা, $y_1={\sec }^2(m{\tan }^{-1}x)\cdot m\times \frac{1}{1+x^2}$

বা, $(1+x^2)y_1={\mathrm{msec}}^2\left({\mathrm{mtan}}^{-1}x\right)$

বা, $(1+x^2)y_1=m\{1+{\tan }^2(m{\tan }^{-1}x)\}$

বা, $(1+x^2)y_1=m(1+y^2)$

বা, $(1+x^2)\cdot y_2+y_1\left(2x\right)=m\left(2y{y}_1\right)$ (প্রমাণিত)