সৃজনশীলঃ ‘f(x)=sin⁡x…’ রাজউক উত্তরা মডেল কলেজ ২০২৪ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$f\left(x\right)=\sin x$

$\therefore f\left(3x\right)=\sin 3x$

এবং $f(\frac{\pi }{2}-3x)=\sin (\frac{\pi }{2}-3x)$

$=\cos 3x$

$\therefore \frac{f\left(3x\right)}{f(\frac{\pi }{2}-3x)}=\frac{\sin 3x}{\cos 3x}=\tan 3x$

মনে করি,

$f\left(x\right)=\tan 3x$

$\therefore f(x+h)=\tan 3(x+h)$

$=\tan (3x+3h)$

মূল নিয়মে অন্তরজের সংজ্ঞানুযায়ী,

$\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$\therefore \frac{d}{dx}(\tan 3x)=\underset{h\to 0}{lim}\frac{\tan (3x+3h)-\tan 3x}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin (3x+3h)}{\cos (3x+3h)}-\frac{\sin 3x}{\cos 3x}}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin (3x+3h)-\cos (3x+3h)\sin 3x}{\cos (3x+3h)\mathrm{cox}3x}}{h}$

$=\underset{h\to 0}{lim}\frac{\sin (3x+3h-3x)}{h\cdot \cos 3x\cdot \cos (3x+3h)}$

$=\underset{h\to 0}{lim}3\times \frac{\sin 3h}{3h\cos 3x\cos (3x+3h)}$

$=3\underset{3h\to 0}{lim}\frac{\sin 3h}{3h}\cdot \underset{h\to 0}{lim}\frac{1}{\cos 3x\cos (3x+3h)}$

$=3\times 1\times \frac{1}{\cos 3x\cos 3x}$

$=\frac{3}{{\cos }^{2}3x}$

$=3{\sec }^{2}3x$

এখানে,

$g\left(x\right)=x+\sqrt{1+x^2}$

এবং

$Z=\left\{g\right(x){\}}^m$

$\therefore Z={(x+\sqrt{1+x^2})}^m$

$\Rightarrow \frac{dz}{dx}=m{(x+\sqrt{1+x^2})}^{m-1}\cdot (1+\frac{2x}{2\sqrt{1+x^2}})$

$\Rightarrow \frac{dz}{dx}=\frac{m{(x+\sqrt{1+x^2})}^m}{(x+\sqrt{1+x^2})}\cdot \frac{(\sqrt{1+x^2}+x)}{\sqrt{1+x^2}}$

$\Rightarrow \frac{\mathrm{dz}}{\mathrm{dx}}=\frac{\mathrm{mz}}{\sqrt{1+x^2}}$

$\Rightarrow \left(\sqrt{1+x^2}\right)\cdot \frac{dz}{dx}=mz$

$\Rightarrow \left(\sqrt{1+x^2}\right){\left(\frac{dz}{dx}\right)}^2=m^2{z}^2$

$\Rightarrow {\left(\frac{\mathrm{dz}}{\mathrm{dx}}\right)}^2+x^2{\left(\frac{\mathrm{dz}}{\mathrm{dx}}\right)}^2=m^2{z}^2$

$\Rightarrow 2\cdot \frac{dz}{dx}\cdot \frac{d^{2}z}{dx}+x^2\{2\cdot \frac{dz}{dx}\cdot \frac{d^{2}z}{dx}\}+{\left(\frac{dz}{dx}\right)}^2\cdot 2x=2m^{2}z\frac{dz}{dx}$

$\Rightarrow \frac{d^{2}z}{dx}+x^2\cdot \frac{d^{2}z}{dx}+x\cdot \frac{dz}{dx}=m^{2}z$

$\Rightarrow (1+x^2)\frac{d^{2}z}{dx}+x\cdot \frac{dz}{dx}-m^{2}z=0$ (Showed)