![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]](/tpl/images/2513/0011/e8c19.png)
![\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}](/tpl/images/2513/0011/1ce8f.png)
![\dfrac{\mathrm d}{\mathrm dx}\left[\sin^{-1}(x)\right] = \dfrac1{\sqrt{1-x^2}}](/tpl/images/2513/0011/f29d0.png)
![\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}](/tpl/images/2513/0011/f6fb8.png)
Find dx/dy.
can you help me in this pleasee.
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Mathematics, 19.10.2021 21:00 Mlmyers81
Find dx/dy.
can you help me in this pleasee.
Answers: 1
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