Question

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

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

Answer 1

The correct option is C $\frac{y}{x}$

Consider the given equation,

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

$\frac{x^2-y^2}{x^2-y^2}=\cos a$

Differentiate with respect to x

$\frac{d}{dx}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\frac{d}{dx}\cos a=0$

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

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

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

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

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

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

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

Hence, this is the answer.