Question

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

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

Answer 1

Answer: (C)

$\frac{y}{x}$

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

$\Rightarrow 2{\tan }^{-1}\frac{y}{x}=\log a\Rightarrow {\tan }^{-1}\frac{y}{x}=\frac{\log a}{2}$

$\Rightarrow \frac{y}{x}=\tan \left(\frac{\log a}{2}\right)$

Now differentiating both sides, we get

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