Practicing Success
Match List-I with List-II. Find the derivatives from List-I.
Choose the correct answer from the options given below : |
(A)-(IV),(B)-(II),(C)-(III),(D)-(I) (A)-(III),(B)-(I),(C)-(IV),(D)-(II) (A)-(I),(B)-(III),(C)-(II),(D)-(IV) (A)-(II),(B)-(IV),(C)-(I),(D)-(III) |
(A)-(II),(B)-(IV),(C)-(I),(D)-(III) |
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) |