(a) y=secx হলে দেখাও যে   $y_{2} = y\left(2 y^{2}-1\right)$

(b) $x^{y} + y^{x} = 0$ সমীকরন হতে $\frac{dy}{dx}$ নির্ণয় কর।

(a) $\mathrm{y}=\sec \mathrm{x} \quad \therefore \mathrm{y}_{1}=\sec \mathrm{x} \tan \mathrm{x}$

$\begin{array}{l} y_{2}=\sec x \cdot \sec ^{2} x+\tan x(\sec x \cdot \tan x)=\sec x\left(\sec ^{2} x+\tan ^{2} x\right) \\ =\sec x\left(2 \sec ^{2} x-1\right)=y\left(2 y^{2}-1\right) \quad(\text { Showed }) \end{array}$

(b) $\mathrm{x}^{\mathrm{y}}+\mathrm{y}^{\mathrm{x}}=0$ সমীকরণ হতে $\frac{\mathrm{dy}}{\mathrm{dx}}$ निর্ণয় কর।

$\mathrm{x}^{\mathrm{y}}+\mathrm{y}^{\mathrm{x}}=0 \quad \therefore \mathrm{x}^{\mathrm{y}}=-\mathrm{y}^{\mathrm{x}} \Rightarrow \mathrm{y} \ln \mathrm{x}=-\mathrm{x} \ln \mathrm{y} \Rightarrow \mathrm{y} \frac{1}{\mathrm{x}}+\ln \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{x} \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}-\ln \mathrm{y}$

$\frac{d y}{d x}\left(\ln x+\frac{x}{y}\right)=-\ln y-\frac{y}{x} \Rightarrow \frac{d y}{d x}=\frac{\frac{-x \ln y-y}{x \ln x+x}}{\underline{x}}=\frac{-y(x \ln y+y)}{x(y \ln x+x)}$