সৃজনশীলঃ ‘f(x)=ln⁡px এবং g(x)=ln⁡x3…’ সিলেট বোর্ড ২০২১ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=\ln px$

$p=3$হলে,

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

এবং$f(x+h)=n3(x+h)$

এখন,

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

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

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

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

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

$=\underset{h\to 0}{lim}\frac{(\frac{h}{x}-\frac{h^2}{2x^2}+\frac{h^3}{3x^2}\cdots \cdots \cdots )}{h}$

$=\underset{h\to 0}{lim}(\frac{1}{x}-\frac{h}{2x^2}+\frac{h^2}{3x^2}\dots \dots \dots )$

$=\frac{1}{x}$

$g\left(x\right)=\ln \sqrt[3]{x}$

$g\left(\frac{1-\cos \phi }{1+\cos \phi }\right)={\ln }^3\sqrt{\frac{1-\cos \phi }{1+\cos \phi }}$

$=\ln {\left(\frac{1-\cos \phi }{1+\cos \phi }\right)}^{\frac{1}{3}}$

$=\frac{1}{3}\ln \left(\frac{1-\cos \phi }{1+\cos \phi }\right)$

$=\frac{1}{3}\ln \frac{2{\sin }^2\frac{\phi }{2}}{2{\cos }^2\frac{\phi }{2}}$

$=\frac{1}{3}\ln ({\tan }^2\frac{\phi }{2})$

$=\frac{1}{3}2\ln \left({\tan }^{\frac{\phi }{2}}\right)$

$\begin{array}{cc}& =\frac{2}{3}\ln (\tan \frac{\phi }{2})\\ & g^{\prime }\frac{(1-\cos \phi )}{(1+\cos \phi )}=\frac{d}{d\phi }\{\frac{2}{3}\ln (\tan \frac{\phi }{2})\}\\ & =\frac{2}{3}\frac{d}{d\theta }\ln (\tan \frac{\phi }{2})\\ & =\frac{2}{3}\frac{1}{\tan \frac{\phi }{2}}\cdot {\sec }^2\frac{\phi }{2}\cdot \frac{1}{2}\\ & =\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{\cos \frac{\phi }{2}}{\sin \frac{\phi }{2}}\cdot \frac{1}{{\cos }^2\frac{\phi }{2}}\\ & =\frac{1}{3}\times \frac{1}{\sin \frac{\phi }{2}\cdot \cos \frac{\phi }{2}}\\ & =\frac{1}{3}\times 2\frac{1}{2\sin \frac{\phi }{2}\cdot \cos \frac{\phi }{2}}\\ & =\frac{2}{3}\times \frac{1}{\sin \phi }\\ & =\frac{2}{3}\mathrm{cosec}\phi \end{array}$