Differentiate both sides of
[tex]y = \dfrac{x\sqrt{a^2-x^2}}2 + \dfrac{a^2}2\sin^{-1}\left(\dfrac xa\right)[/tex]
with respect to y ; by the product rule,
[tex]1 = \dfrac12\sqrt{a^2-x^2}\dfrac{\mathrm dx}{\mathrm dy} + \dfrac x2 \dfrac{\mathrm d}{\mathrm dy}\left[\sqrt{a^2-x^2}\right] + \dfrac{a^2}2\dfrac{\mathrm d}{\mathrm dy}\left[\sin^{-1}\left(\dfrac xa\right)\right][/tex]
Use the chain rule for the remaining derivatives.
[tex]\dfrac{\mathrm d}{\mathrm dy}\left[\sqrt{a^2-x^2}\right] = \dfrac1{2\sqrt{a^2-x^2}}\dfrac{\mathrm d}{\mathrm dy}\left[a^2-x^2\right] \\\\ = \dfrac{-2x}{2\sqrt{a^2-x^2}}\dfrac{\mathrm dx}{\mathrm dy} \\\\ = -\dfrac x{\sqrt{a^2-x^2}} \dfrac{\mathrm dx}{\mathrm dy}[/tex]
Recall that
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\sin^{-1}(x)\right] = \dfrac1{\sqrt{1-x^2}}[/tex]
Then
[tex]\dfrac{\mathrm d}{\mathrm dy}\left[\sin^{-1}\left(\dfrac xa\right)\right] = \dfrac1{\sqrt{1-\left(\frac xa\right)^2}}\dfrac{\mathrm d}{\mathrm dy}\left[\dfrac xa\right] \\\\ = \dfrac1{a\sqrt{1-\left(\frac xa\right)^2}}\dfrac{\mathrm dx}{\mathrm dy} \\\\ = \dfrac1{\sqrt{a^2}\sqrt{1-\left(\frac xa\right)^2}}\dfrac{\mathrm dx}{\mathrm dy} \\\\ = \dfrac1{\sqrt{a^2-x^2}}\dfrac{\mathrm dx}{\mathrm dy}[/tex]
Putting everything together, we have
[tex]1 = \dfrac{\sqrt{a^2-x^2}}2\dfrac{\mathrm dx}{\mathrm dy} - \dfrac{x^2}{2\sqrt{a^2-x^2}}\dfrac{\mathrm dx}{\mathrm dy} + \dfrac{a^2}{2\sqrt{a^2-x^2}}\dfrac{\mathrm dx}{\mathrm dy}[/tex]
[tex]1 = \left(\dfrac{\sqrt{a^2-x^2}}2+\dfrac{a^2-x^2}{2\sqrt{a^2-x^2}}\right)\dfrac{\mathrm dx}{\mathrm dy}[/tex]
[tex]1 = \dfrac1{2\sqrt{a^2-x^2}}\bigg((a^2-x^2) + (a^2-x^2)\bigg)\dfrac{\mathrm dx}{\mathrm dy}[/tex]
[tex]1 = \dfrac{2a^2-2x^2}{2\sqrt{a^2-x^2}}\dfrac{\mathrm dx}{\mathrm dy}[/tex]
[tex]1 = \sqrt{a^2-x^2}\dfrac{\mathrm dx}{\mathrm dy}[/tex]
[tex]\boxed{\dfrac{\mathrm dx}{\mathrm dy} = \dfrac1{\sqrt{a^2-x^2}}}[/tex]
