Implicit differentiation of In(x^2 + y^2) = 2arcsin(x/y)

[chengeto]

Find $\displaystyle \frac{dy}{dx}$ at an arbtrary point on this curve $\displaystyle ln(x^2+y^2)\, =\,2arcsin \,\frac{x}{y}$

My attempt to solution:

$\displaystyle \frac{1}{x^2+y^2}2x+(4y^2)(\frac{dy}{dx})=2\frac{1}{\sqrt1-(\frac{x}{y})^2}\frac{y-x\frac{dy}{dx}}{y^2}$

Here l get stuck with how to isolate for dy/dx. I also do not know if l differentiated $\displaystyle \frac{x}{y}$ correctly.

 

[soroban]

Hello, chengeto!

This is The Implicit Differentiation Problem from **** (coming to a theater near you).

$\displaystyle \text{Find }\frac{dy}{dx}\,\text{ for: }\:\ln(x^2+y^2)\: =\:2\arcsin\left(\tfrac{x}{y}\right)$

. . . . . . . $\displaystyle \frac{1}{x^2+y^2}\cdot\left(2x + 2y\,\frac{dy}{dx}\right) \;=\;2\cdot\frac{1}{\sqrt{1 - (\frac{x}{y})^2}}\cdot\frac{y-x\,\frac{dy}{dx}}{y^2}$

. . . . . . . . . . . . . . $\displaystyle \frac{2\left(x + y\,\frac{dy}{dx}\right)}{x^2+y^2} \;=\;\frac{2\left(y - x\,\frac{dy}{dx}\right)}{\sqrt{\frac{y^2-x^2}{y^2}}\cdot y^2}$

. . . . . . . . . . . . . . . . .$\displaystyle \frac{x + y\,\frac{dy}{dx}}{x^2+y^2} \;=\;\frac{y - x\,\frac{dy}{dx}}{y\sqrt{y^2-x^2}}$

. . . . . .$\displaystyle y\sqrt{y^2-x^2}\cdot\left(x + y\,\frac{dy}{dx}\right) \;=\;(x^2+y^2)\cdot\left(y - x\,\frac{dy}{dx}\right)$

. .$\displaystyle xy\sqrt{y^2-x^2} + y^2\sqrt{y^2-x^2}\,\frac{dy}{dx} \;=\;y(x^2+y^2) - x(x^2+y^2)\,\frac{dy}{dx}$

$\displaystyle y^2\sqrt{y^2-x^2}\,\frac{dy}{dx} + x(x^2+y^2)\,\frac{dy}{dx} \;=\;y(x^2+y^2) - xy\sqrt{y^2-x^2}$

. $\displaystyle \bigg[y^2\sqrt{y^2-x^2} + x(x^2+y^2)\bigg]\,\frac{dy}{dx} \;=\;y\bigg[x^2+y^2 - x\sqrt{y^2-x^2}\bigg]$

. . . . . . . . . . . . . . . . . . . $\displaystyle \frac{dy}{dx} \;=\;\frac{y(x^2+y^2-x\sqrt{y^2-x^2})}{y^2\sqrt{y^2-x^2} + x(x^2+y^2)}$

 

[chengeto]

Soroban thanks very much for the help.