(a) মান নির্ণয় কর : dy/dx, যখন   $y = \cot^{- 1}{\left \lbrace \left ( \sqrt{1 + x^{2}} - x \right ) \right \rbrace}$

(b) মান নির্ণয় কর : dy/dx, যখন  $x^{a} y^{a} = \left ( x - y \right )^{a + b}$  

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

ধরি, $x=\tan \theta$

$\begin{array}{l} y=\cot ^{-1}(\sec \theta-\tan \theta)=\cot ^{-1}\left\{\frac{1-\sin \theta}{\cos \theta}\right\}=\cot ^{-1}\left\{\frac{\sin ^{2} \frac{\theta}{2}-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}{\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}}\right\} \\ =\cot ^{-1}\left\{\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}\right\}=\cot ^{-1}\left\{\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}\right\}=\cot ^{-1}\left\{\frac{\tan \frac{\pi}{4}-\tan \frac{\theta}{2}}{1+\tan \frac{\pi}{4} \tan \frac{\theta}{2}}\right\}=\cot ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right\} \\ =\cot ^{-1}\left\{\cot \left\{\frac{\pi}{2}-\frac{\pi}{4}+\frac{\theta}{2}\right\}\right\}=\frac{\pi}{4}+\frac{\theta}{2}=\frac{\pi}{4}+\frac{\tan ^{-1} x}{2} \therefore \frac{d y}{d x}=\frac{1}{2} \times \frac{1}{1+x^{2}}=\frac{1}{2\left(1+x^{2}\right)} \text { (Ans.) } \end{array}$

(b) dy/dx, যখন $x^{a} y^{b}=(x-y)^{a+b}$

$\begin{array}{l} x^{a} y^{b}=(x-y)^{a+b} \Rightarrow \ln \left(x^{a} \cdot y^{b}\right)=\ln (x-y)^{a+b} \Rightarrow \ln x^{a}+\ln y^{b}=(a+b) \ln (x-y) \\ \Rightarrow a \ln x+b \ln y=(a+b) \ln (x-y) \Rightarrow \frac{a}{x}+\frac{b}{y} \cdot \frac{d y}{d x}=\frac{a+b}{x-y}\left(1-\frac{d y}{d x}\right) \Rightarrow \frac{a}{x}+\frac{b}{y} \cdot \frac{d y}{d x}=\frac{a+b}{x-y}-\frac{a+b}{x-y} \cdot \frac{d y}{d x} \\ \Rightarrow \frac{b}{y} \cdot \frac{d y}{d x}+\frac{a+b}{x-y} \cdot \frac{d y}{d x}=\frac{a+b}{x-y}-\frac{a}{x} \Rightarrow\left(\frac{b}{y}+\frac{a+b}{x-y}\right) \frac{d y}{d x}=\frac{a x+b x-a x+a y}{x(x-y)} \\ \Rightarrow\left\{\frac{b x-b y+a y+b y}{y(x-y)}\right\} \frac{d y}{d x}=\frac{b x+a y}{x(x-y)} \Rightarrow \frac{d y}{d x}=\frac{y}{x} \cdot \frac{b x+a y}{b x+a y}[x \neq y] \therefore \frac{d y}{d x}=\frac{y}{x} \end{array}$