Question

help................... ​

Answer 1

$\underline{\text{Proof :}}$

$\mathrm{Given,\:x+\sqrt{xy}+y=1}$

$\text{Diff. both sides w. r. to x, we get}$

$\mathrm{\frac{d}{dx}(x+\sqrt{xy}+y)=\frac{d}{dx}(1)}$

$\to \mathrm{\frac{dx}{dx}+\frac{d}{dx}\sqrt{xy}+\frac{dy}{dx}=0}$

$\to \mathrm{1 + \frac{d}{dx}(\sqrt{x}\:\sqrt{y})+\frac{dy}{dx}=0}$

$\to \small{\mathrm{1 +\sqrt{x}\frac{d}{dx}(\sqrt{y})+\sqrt{y}\frac{d}{dx}(\sqrt{x})+\frac{dy}{dx}=0}}$

$\to \mathrm{1+\frac{\sqrt{x}}{2\sqrt{y}}\frac{dy}{dx}+\frac{\sqrt{y}}{2\sqrt{x}}+\frac{dy}{dx}=0}$

$\to \mathrm{(\frac{\sqrt{x}}{2\sqrt{y}}+1)\frac{dy}{dx}=-(1+\frac{\sqrt{y}}{2\sqrt{x}})}$

$\to \mathrm{\frac{\sqrt{x}+2\sqrt{y}}{2\sqrt{y}}\frac{dy}{dx}=-\frac{2\sqrt{x}+\sqrt{y}}{2\sqrt{x}}}$

$\to \mathrm{\frac{\sqrt{x}+2\sqrt{y}}{\sqrt{y}}\frac{dy}{dx}=-\frac{2\sqrt{x}+\sqrt{y}}{\sqrt{x}}}$

$\to \boxed{\mathrm{\frac{dy}{dx}=-\frac{\sqrt{y}(2\sqrt{x}+\sqrt{y})}{\sqrt{x}(\sqrt{x}+2\sqrt{y})}}}$

$\text{Hence, proved.}$

$\underline{\text{Rule :}}$

$\mathrm{1.\:\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}}$

$\mathrm{2.\:\frac{d}{dx}\sqrt{y}=\frac{1}{2\sqrt{y}}\frac{dy}{dx}}$

$\mathrm{3.\:\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}}$