Question

If `x^2 y^2 = sin^-1 sqrt(x^2 + y^2) + cos^-1 sqrt(x^2 + y^2)`, then `"dy"/"dx"` = ?

Options

• `(- x)/y`

• `x/y`

• `(- y)/x`

• `(y)/x`

Answer 1

`(- y)/x`

Explanation:

We have,

`x^2 y^2 = sin^-1 sqrt(x^2 + y^2) + cos^-1 sqrt(x^2 + y^2)`

`=> x^2y^2 = pi/2`

On differentiating w.r.t.,'x' we get

`2x^2y "dy"/"dx" + 2xy^2 = 0`

`"dy"/"dx" = (- 2"xy"^2)/(2x^2y) = - y/x`