Q.If $x^y=y^x$, then $\dfrac{dy}{dx}=$ (A) $\dfrac{x(x\log y-y)}{y(y\log x-x)}$ (B) $\dfrac{y(y\log x-x)}{x(x\log y-y)}$ (C) $\dfrac{y^2(1-\log x)}{x^2(1-\log y)}$ (D) $\dfrac{y(1-\log x)}{x(1-\log y)}$

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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