সৃজনশীলঃ ‘F(x)=cot⁡5x এবং y=cos⁡(psin−1⁡x).…’ আদমজী ক্যান্টনমেন্ট কলেজ ২০২৬ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

মনে করি, $y=F\left(x\right)=\cot 5x$

$\therefore F(x+h)=\cot 5(x+h)$

বা, $F(x+h)=\cot (5x+5h)$

অন্তরজের সংজ্ঞানুসারে, $\frac{dy}{dx}=\underset{h\to 0}{lim}\frac{F(x+h)-F\left(x\right)}{h}$

$\therefore \frac{d}{dx}(\cot 5x)=\underset{h\to 0}{lim}\frac{\cot (5x+5h)-\cot 5x}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{\cos (5x+5h)}{\sin (5x+5h)}-\frac{\cos 5x}{\sin 5x}}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin 5x\cos (5x+5h)-\cos 5x\cdot \sin (5x+5h)}{\sin (5x+5h)\sin 5x}}{h}$

$=\underset{h\to 0}{lim}\frac{\sin 5x\cdot \cos (5x+5h)-\cos 5x\cdot \sin (5x+5h)}{h\cdot \sin (5x+5h)\sin 5x}$

$=\underset{h\to 0}{lim}\frac{\sin (5x-5x-5h)}{h\sin (5x-5h)\sin 5x}$

$=\underset{h\to 0}{lim}\frac{-\sin 5h}{h}\underset{h\to 0}{lim}\frac{1}{\sin (5x+5h)\sin 5x}$

$=\underset{h\to 0}{lim}5\frac{\sin 5h}{5h}\underset{h\to 0}{lim}\frac{1}{\sin (5x+5h)\sin 5x}$

$=-5.1\frac{1}{\sin (5x+0)\sin 5x}$

$=\frac{-5}{{\sin }^{2}5x}=-5{\mathrm{cosec}}^{2}5x$

$\therefore \frac{d}{dx}(\cot 5x)=-5{\mathrm{cosec}}^{2}5x.$

দেওয়া আছে, $y=\cos (p{\sin }^{-1}x)$

বা, ${\cos }^{-1}y=p{\sin }^{-1}x$

বা, $\frac{d}{dx}({\cos }^{-1}y)=\frac{d}{dx}(p{\sin }^{-1}x)$

বা, $\frac{-y_1}{\sqrt{1-y^2}}=\frac{p}{\sqrt{1-x^2}}$

বা, $\frac{y^2}{1-y^2}=\frac{p^2}{1-x^2}$ [বর্গ করে]

বা, $p^2(1-y^2)=(1-x^2)y_1{}^2$

বা, $\frac{d}{dx}\left\{p^2(1-y^2)\right\}=\frac{d}{dx}(1-x^2)\cdot y_1{}^2$

বা, $p^2(-2y{y}_1)=(-2x)y_1{}^2+(1-x^2)\cdot \left(2y{y}_2\right)$

বা, $-p^{2}y=-x{y}_1+(1-x^2)y_2$ [উভয়পক্ষে $2y_1$ দ্বারা ভাগ করে]

$\therefore (1-x^2)y_2-{\mathrm{xy}}_1+p^{2}y=0$. (দেখানো হলো)