Question

If ${\cos }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ (a constant), then $\frac{dy}{dx}=$

• A. $\frac{y}{x}$
• B. $\frac{x}{y}$
• C. $\frac{x^2}{y^2}$
• D. $\frac{y^2}{x^2}$

Answer 1

Answer: (A)

$\frac{y}{x}$

${\cos }^{-1}\frac{x^2-y^2}{x^2+y^2}=k$

$\frac{x^2-y^2}{x^2+y^2}=cos\left(k\right)$

On differentiating both sides, we get

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

$(2x^3+2x{y}^2-2x^{2}y\frac{dy}{dx}-2y^3\frac{dy}{dx})-(2x^3-2x{y}^2+2x^{2}y\frac{dy}{dx}-2y^3\frac{dy}{dx})=0$

$4x{y}^2-4y{x}^2\frac{dy}{dx}=0$

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