Question

If $\sqrt{y+x}+\sqrt{y-x}=c,\mathrm{show}\mathrm{that}\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}$

Answer 1

$$\begin{gathered} Here, \\ \sqrt{y+x}+\sqrt{y-x}=c \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(\sqrt{y+x}\right)+\frac{d}{dx}\sqrt{y-x}=\frac{d}{dx}\left(c\right) \\ \Rightarrow \frac{1}{2\sqrt{y+x}}\frac{d}{dx}\left(y+x\right)+\frac{1}{2\sqrt{y-x}}\frac{d}{dx}\left(y-x\right)=0 \\ \Rightarrow \frac{1}{2\sqrt{y+x}}\left(\frac{dy}{dx}+1\right)+\frac{1}{2\sqrt{y-x}}\left(\frac{dy}{dx}-1\right)=0 \\ \Rightarrow \frac{dy}{dx}\left(\frac{1}{2\sqrt{y+x}}\right)+\frac{dy}{dx}\left(\frac{1}{2\sqrt{y-x}}\right)=\frac{1}{2\sqrt{y-x}}-\frac{1}{2\sqrt{y+x}} \\ \Rightarrow \frac{dy}{dx}\times \frac{1}{2}\left[\frac{1}{\sqrt{y+x}}+\frac{1}{\sqrt{y-x}}\right]=\frac{1}{2}\left[\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y-x}\sqrt{y+x}}\right] \\ \Rightarrow \frac{dy}{dx}\left[\frac{\sqrt{y-x}+\sqrt{y+x}}{\sqrt{y+x}\sqrt{y-x}}\right]=\left[\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y-x}\sqrt{y+x}}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}}\times \frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}-\sqrt{y-x}}\left[\mathrm{rationalizing}\mathrm{the}\mathrm{denominator}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{\left(y+x\right)+\left(y-x\right)-2\sqrt{y+x}\sqrt{y-x}}{y+x-y+x} \\ \Rightarrow \frac{dy}{dx}=\frac{2y-2\sqrt{y^2-x^2}}{2x} \\ \Rightarrow \frac{dy}{dx}=\frac{2y}{2x}-\frac{2\sqrt{y^2-x^2}}{2x} \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2-x^2}{x^2}} \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1} \end{gathered}$$