Question

If ${\sin }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$, then $\frac{dy}{dx}$ is equal to?

Answer 1

We have,

${\sin }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$

$\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin \left(\log a\right)$ …….. (1)

On differentiating w.r.t $x$, we get

$\frac{(x^2+y^2)(2x-2y\frac{dy}{dx})-(x^2-y^2)(2x+2y\frac{dy}{dx})}{{(x^2+y^2)}^2}=0$

$(x^2+y^2)(2x-2y\frac{dy}{dx})-(x^2-y^2)(2x+2y\frac{dy}{dx})=0$

$(x^2+y^2)(2x-2y\frac{dy}{dx})=(x^2-y^2)(2x+2y\frac{dy}{dx})$

$2x(x^2+y^2)-2y\frac{dy}{dx}(x^2+y^2)=2x(x^2-y^2)+2y\frac{dy}{dx}(x^2-y^2)$

$2x^3+2x{y}^2-2x^{2}y\frac{dy}{dx}-2y^3\frac{dy}{dx}=2x^3-2x{y}^2+2x^{2}y\frac{dy}{dx}-2y^3\frac{dy}{dx}$

$2x{y}^2-2x^{2}y\frac{dy}{dx}=-2x{y}^2+2x^{2}y\frac{dy}{dx}$

$4x{y}^2=4x^{2}y\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{y}{x}$

Hence, this is the answer.