If log sqrt{x^2+y^2}=tan ^{-1}left(frac{x}{y} | vedclass

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Question

(A) Given equation: $\log \sqrt{x^2+y^2}=	an ^{-1}\left(\frac{x}{y}ight)$ Differentiating both sides with respect to $x$: $\frac{1}{\sqrt{x^2+y^2}}

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$\frac{y-x}{y+x}$

Vedclass · Answer

(A) Given equation: $\log \sqrt{x^2+y^2}=	an ^{-1}\left(\frac{x}{y}ight)$ Differentiating both sides with respect to $x$: $\frac{1}{\sqrt{x^2+y^2}} \cdot \frac{d}{dx}(\sqrt{x^2+y^2}) = \frac{1}{1+(\frac{x}{y})^2} \cdot \frac{d}{dx}(\frac{x}{y})$ $\frac{1}{\sqrt{x^2+y^2}} \cdot \frac{1}{2\sqrt{x^2+y^2}} \cdot (2x + 2y \frac{dy}{dx}) = \frac{y^2}{x^2+y^2} \cdot \frac{y - x \frac{dy}{dx}}{y^2}$ $\frac{2(x + y \frac{dy}{dx})}{2(x^2+y^2)} = \frac{y - x \frac{dy}{dx}}{x^2+y^2}$ $x + y \frac{dy}{dx} = y - x \frac{dy}{dx}$ $y \frac{dy}{dx} + x \frac{dy}{dx} = y - x$ $\frac{dy}{dx}(x + y) = y - x$ $\frac{dy}{dx} = \frac{y-x}{x+y}$ Thus,the correct option is $A$.

If $\log \sqrt{x^2+y^2}=\tan ^{-1}\left(\frac{x}{y}\right)$,then $\frac{d y}{d x}$ is equal to

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