Match List-I with List-II. List-I List-II A xy +x + y = 1 I y'(1, e)=…

The correct answer is Option (3) → A-III, B-IV, C-I, D-II

(A) $\frac{d(xy+x+y)}{dx}=x\frac{dy}{dx}y+1+\frac{dy}{dx}$

$=(x+1)\frac{dy}{dx}+(y+1)$

$\frac{dy}{dx}=-\frac{(y+1)}{(x+1)}$

$y'(1,0)=-\frac{(0+1)}{(1+1)}=\frac{1}{2}$

(B) $\frac{d(x^x+y)}{dx}=0$

$x^x(ln\,x+1)+\frac{dy}{dx}=0$

$⇒\frac{dy}{dx}=-x^x(ln\,x+1)$

$y'(1,1)=-1'(ln\,1+1)=-1$

(C) $\frac{d(x^2+x\log y)}{dx}=2x+\log y+(\frac{x}{y})\frac{dy}{dx}$

$⇒\frac{dy}{dx}=\frac{-y}{x}×(2x+\log y)$

$⇒f'(1,e)=-\frac{e}{1}(2+1)=-3e$

(D) $\frac{d(x^3y^2)}{dx}=3x^2y^2+2x^3y\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{x^2y^2}{x^3y}$

$f'(2,2)=-\frac{3}{2}$