সৃজনশীলঃ ‘h(x)=cos⁡3x,u=tan−1⁡2x…’ রাজশাহী বোর্ড ২০২২ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, h(x) = cos 3x

মূল নিয়মের সংগানুসারে, 

$\begin{array}{cc}& h^{\prime }\left(x\right)=\frac{d}{dx}h\left(x\right)=\underset{h\to 0}{lim}\frac{\cos 3(x+h)-\cos 3x}{h}\\ & =\underset{h\to 0}{lim}\frac{1}{h}\{\cos (3x+3h)-\cos 3x\}\\ & =\underset{h\to 0}{lim}\frac{1}{h}2\cdot \sin \frac{3x+3h+3x}{2}\sin \frac{3x-3x-3h}{2}\\ & =\underset{h\to 0}{lim}\frac{2}{h}\sin \frac{6x+3h}{2}\sin \frac{-3h}{2}\\ & =\underset{h\to 0}{lim}\sin \frac{6x+3h}{2}\times \underset{h\to 0}{lim}\frac{\sin \left(\frac{-3h}{2}\right)}{\left(\frac{-3h}{2}\right)}\times \frac{-3h}{2}\times 2\\ & =\sin \frac{6x+0}{2}\times 1\times \frac{-3}{2}\times 2\\ & =-3\sin 3x\left(\text{ Ans }\right)\end{array}$

দেওয়া আছে, $u={\tan }^{-1}\left(2x\right)\Rightarrow \tan u=2x..........\left(i\right)$

এবং, $\ln y=u$

$\begin{array}{cc}& \Rightarrow \ln y={\tan }^{-1}2x\\ & \Rightarrow \frac{1}{y}\cdot y_1=\frac{1\times 2}{1+4x^2}\\ & \Rightarrow y_1(1+4x^2)=2y\\ & \Rightarrow y_1(0+8x)+(1+4x+y_2=2y\end{array}$

(i) নং থেকে $\tan u=2x$

$\begin{array}{cc}& \Rightarrow {\tan }^{2}u=4x\\ & \Rightarrow 1+{\tan }^{2}u=1+4x^2\\ & \Rightarrow {\sec }^{2}u=(1+4x^2)\cdot \cdot \left(ii\right)\end{array}$

এখন, (ii) নং থেকে,

$y_1\times 8x+(1+4x^2)y_2=2y_1$

$\Rightarrow y_1\times 8x+(1+4x^2)y_2-2y_1=0$

$\Rightarrow 4y_1\tan u+{\sec }^2\left(u\right)y_2-2y_1=0$

${\sec }^{2}u\cdot y_2+2y_1(2\tan u-1)=0$ (Proved)