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

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Question

(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,$\frac{dy}{dx} = -\frac{198x}{202y} = -\frac{99x}{101y}$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

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

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