Q.If $x\sqrt{y+1}+y\sqrt{x+1}=0$ and $x\ne y$ then $\dfrac{dy}{dx}=$ (A) $\dfrac{1}{(1+x)^2}$ (B) $-\dfrac{1}{(1+x)^2}$ (C) $(1+x)^2$ (D) $-\dfrac{x}{x+1}$

Differentiate both sides of $\sqrt x+\sqrt y=\sqrt a$ implicitly with respect to $x$.

$\sqrt x+\sqrt y=\sqrt a$

Differentiating w.r.t. $x$ (note $\sqrt a$ is a constant):

$\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\cdot\frac{dy}{dx}=0$ …

Differentiate $\log(x+y)=2xy$ implicitly and evaluate at $x=0$ (where $y=1$).

At $x=0$: $\log(0+y)=0\Rightarrow y=1$.

Differentiating both sides w.r.t. $x$: $\dfrac{1+y'}{x+y}=2(y+xy')$.

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Differentiate implicitly using the product rule on both terms.

$x\sin y + y\sin x = 0$

Differentiating both sides w.r.t. $x$:

$\sin y + x\cos y\cdot\dfrac{dy}{dx} + \dfrac{dy}{dx}\sin x + y\cos x = 0$

$\dfrac{dy}{dx}\big(x\cos y + \sin x\big) = -(\sin y + y\cos x)$

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