Question

Find dx/dy.
can you help me in this pleasee....... ​

Answer 1

( ((a^2 - x^2)^1/2 + (1/2(a-x^2)^(-1/2)*2a-2x)*(2x)) / 4 ) + ((4a/4)*sin^-1(x-a)) + ((a^2/2)*((a-x)/a^2*sqrt(1-(x-a^2)))

Answer 2

Differentiate both sides of

`y = \frac{x√(a^2-x^2)}2 + \frac{a^2}2\sin^(-1)\left(\frac xa\right)`

with respect to y ; by the product rule,

`1 = \frac12√(a^2-x^2)(\mathrm dx)/(\mathrm dy) + \frac x2 (\mathrm d)/(\mathrm dy)\left[√(a^2-x^2)\right] + \frac{a^2}2(\mathrm d)/(\mathrm dy)\left[\sin^(-1)\left(\frac xa\right)\right]`

Use the chain rule for the remaining derivatives.

`(\mathrm d)/(\mathrm dy)\left[√(a^2-x^2)\right] = \frac1{2√(a^2-x^2)}(\mathrm d)/(\mathrm dy)\left[a^2-x^2\right] \\\\ = (-2x)/(2√(a^2-x^2))(\mathrm dx)/(\mathrm dy) \\\\ = -\frac x{√(a^2-x^2)} (\mathrm dx)/(\mathrm dy)`

Recall that

`(\mathrm d)/(\mathrm dx)\left[\sin^(-1)(x)\right] = \frac1{√(1-x^2)}`

Then

`(\mathrm d)/(\mathrm dy)\left[\sin^(-1)\left(\frac xa\right)\right] = \frac1{√(1-\left(\frac xa\right)^2)}(\mathrm d)/(\mathrm dy)\left[\frac xa\right] \\\\ = \frac1{a√(1-\left(\frac xa\right)^2)}(\mathrm dx)/(\mathrm dy) \\\\ = \frac1{√(a^2)√(1-\left(\frac xa\right)^2)}(\mathrm dx)/(\mathrm dy) \\\\ = \frac1{√(a^2-x^2)}(\mathrm dx)/(\mathrm dy)`

Putting everything together, we have

`1 = \frac{√(a^2-x^2)}2(\mathrm dx)/(\mathrm dy) - (x^2)/(2√(a^2-x^2))(\mathrm dx)/(\mathrm dy) + (a^2)/(2√(a^2-x^2))(\mathrm dx)/(\mathrm dy)`

`1 = \left(\frac{√(a^2-x^2)}2+(a^2-x^2)/(2√(a^2-x^2))\right)(\mathrm dx)/(\mathrm dy)`

`1 = \frac1{2√(a^2-x^2)}\bigg((a^2-x^2) + (a^2-x^2)\bigg)(\mathrm dx)/(\mathrm dy)`

`1 = (2a^2-2x^2)/(2√(a^2-x^2))(\mathrm dx)/(\mathrm dy)`

`1 = √(a^2-x^2)(\mathrm dx)/(\mathrm dy)`

`\boxed{(\mathrm dx)/(\mathrm dy) = \frac1{√(a^2-x^2)}}`