Match List-I with List-II. Find the derivatives from List-I. List-I L…

The correct answer is Option (4) → (A)-(II),(B)-(IV),(C)-(I),(D)-(III)

(A) $y=\sqrt{\sin x+y}$

so $y^2=\sin x+y$

differentiating wrt (x)

$2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}$

so $\frac{dy}{dx}=\frac{\cos x}{2y-1}$ (II)

(B) $\sin y= x\sin(a+y)$

$x=\frac{\sin y}{\sin(a+y)}$ differentiating wrt (y)

$\frac{dx}{dy}=\frac{(\cos y\sin(a+y)-\cos(a+y)\sin y}{\sin^2(a+y)}=\frac{\sin(y)}{\sin^2(a+y)}$

so $\frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin(y)}$ (IV)

(C) $y+\sin y=\cos x$

differentiating wrt (x)

$\frac{dy}{dx}=(1+\cos y)=-\sin x⇒\frac{dy}{dx}=\frac{-\sin x}{1+\cos y}$ (I)

(D) $y=\tan(x+y)$  ...(1)

so $\tan^{-1}y=x+y$

differentiating wrt y

$\frac{1}{1+y^2}=\frac{dx}{dy}+1$ so $\frac{dy}{dx}=-\frac{(1+y^2)}{y^2}$

$⇒\frac{dy}{dx}=-\frac{(1+\tan^2(x+y))}{\tan^2(x+y)}$

$=\frac{-\sec^2(x+y)}{\tan^2(x+y)}$

$=\frac{-1}{\sin^2(x+y)}$ (III)