How do i find the derivative dy/dx of given curve. (y is fn of x)

Question

the curve is $\dfrac{x+\sqrt{(a^2-y^2)}}{a}$ = $\log\dfrac{(a+\sqrt{(a^2+y^2)}}{y}$ i need to find the length of tangent for which i need to evaluate slope $\dfrac{dy}{dx}$. please help

Answer 1

Differentiate implicitly the given equation:

$$\frac1adx-\frac{y}{a\sqrt{a^2-y^2}}dy=\frac y{a+\sqrt{a^2+y^2}}\cdot\frac{\frac{y^2}{\sqrt{a^2+y^2}}-a-\sqrt{a^2+y^2}}{y^2}dy\iff$$

$$\color{red}{\frac1adx-\frac{y}{a\sqrt{a^2-y^2}}dy}=\frac y{a+\sqrt{a^2+y^2}}\cdot\frac{-a\sqrt{a^2+y^2}-a^2}{y^2}dy=\color{red}{-\frac ay dy}$$

and now do $\;\cfrac{dy}{dx}\;$ and solve...