সৃজনশীলঃ ‘f(x)=log5⁡x,g(x)=sec⁡x…’ যশোর বোর্ড ২০২১ – অন্তরীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$f\left(x\right)={\log }_{5}x={\log }_{5}e\times {\log }_{e}x$

$={\log }_{5}e\times \ln x$

$\therefore f(x+h)={\log }_{5}ex\ln (x+h)$

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

$\therefore \frac{d}{dx}({\log }_{5}x)=\underset{h\to 0}{lim}\frac{{\log }_{5}ex\ln (x+h)-{\log }_{5}e\times \ln x}{h}$

$=\underset{h\to 0}{lim}{\log }_5{e}^{\frac{\ln (x+h)-\ln x}{h}}$

$={\log }_5{e}_{h\to 0}\frac{\ln \left(\frac{x+h}{x}\right)}{h}$

$={\log }_5{e}_{h\to 0}\frac{\ln (1+\frac{h}{x})}{h}$

$={\log }_5{\lim }_{h\to 0}\frac{1}{h}(\frac{h}{x}-\frac{1}{2}\frac{h^2}{x^2}+\frac{1}{3}\cdot \frac{h^3}{x^3}\dots \dots .)$

$={\log }_5{e}_{h\to 0}(\frac{1}{x}-\frac{h}{2x}+\frac{h^2}{3x^2}\dots \dots )$

$={\log }_5{e}^{\frac{1}{x}}$

$\frac{d}{dx}({\log }_{5}x)=\frac{1}{x}{\log }_{5}e$

$g\left(x\right)=\sec x$

এবং $\frac{1}{g\left(\sqrt{y}\right)}=2x$

$\Rightarrow \frac{1}{\sec \sqrt{y}}=2x$

$\Rightarrow \cos \sqrt{y}=2x$

$\Rightarrow \sqrt{y}={\cos }^{-1}\left(2x\right)$

$\Rightarrow y={({\cos }^{-1}2x)}^2$

x এর সাপেক্ষে অন্তরকরন করে পাই,

$\Rightarrow y_1=2({\cos }^{-1}2x)\cdot \frac{1}{\sqrt{1-(2x{)}^2}}\cdot —2$

$\Rightarrow y_1=\frac{-4{\cos }^{-1}2x}{\sqrt{1-4x^2}}$

$\Rightarrow y_1^2=\frac{16{({\cos }^{-1}2x)}^2}{1-4x^2}$

$\Rightarrow y_1^2(1-4x^2)=16y$

$\Rightarrow \frac{d}{\mathrm{dx}}\left\{(1-4x^2)y_1{}^2\right\}=\frac{d}{\mathrm{dx}}\left(16y\right)$

$\Rightarrow (1-4x^2)\cdot 2y_1\cdot y_2+y_1{}^2(-8x)=16y_1$

$\Rightarrow (1-4x^2)y_2-4x{y}_1-8=0$ (Showed)