If log _{10}left(frac{x^2-y^2}{x^2+y^2}right) | vedclass

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(A) Given,$\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$. Applying the definition of logarithm,we get $\frac{x^2-y^2}{x^2+y^2} = 10^2 = 100$. Cros

Vedclass · Accepted Answer

$-\frac{99x}{101y}$

Vedclass · Answer

(A) Given,$\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$. Applying the definition of logarithm,we get $\frac{x^2-y^2}{x^2+y^2} = 10^2 = 100$. Cross-multiplying,we have $x^2 - y^2 = 100(x^2 + y^2)$. $x^2 - y^2 = 100x^2 + 100y^2$. Rearranging the terms,we get $x^2 - 100x^2 = 100y^2 + y^2$,which simplifies to $-99x^2 = 101y^2$. Now,differentiate both sides with respect to $x$: $\frac{d}{dx}(-99x^2) = \frac{d}{dx}(101y^2)$. $-99(2x) = 101(2y) \frac{dy}{dx}$. $-198x = 202y \frac{dy}{dx}$. Therefore,$\frac{dy}{dx} = -\frac{198x}{202y} = -\frac{99x}{101y}$.

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

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