Question

If ${\log }_{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$, then $\frac{dy}{dx}=$............

• A. $-\frac{99x}{101y}$
• B. $\frac{99x}{101y}$
• C. $-\frac{99y}{101x}$
• D. $\frac{99y}{101x}$

Answer 1

The correct option is A $-\frac{99x}{101y}$

${\log }_{10}(x^2-y^2/x^2+y^2)=2\Rightarrow \frac{x^2-y^2}{x^2+y^2}={10}^2\Rightarrow 99x^2=-101y^2$

Differentiating both the sides

$\Rightarrow \frac{dy}{dx}=-\frac{99x}{101y}$