If xy = tan^{-1}(xy) + cot^{-1}(xy),then frac | vedclass

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(B) We know that for any real number $z$,$	an^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$.Given the equation $xy = 	an^{-1}(xy) + \cot^{-1}(xy)$,we can

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$-\frac{y}{x}$

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(B) We know that for any real number $z$,$	an^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$.Given the equation $xy = 	an^{-1}(xy) + \cot^{-1}(xy)$,we can substitute $z = xy$ to get $xy = \frac{\pi}{2}$.Now,differentiate both sides with respect to $x$ using the product rule:$\frac{d}{dx}(xy) = \frac{d}{dx}(\frac{\pi}{2})$$x \frac{dy}{dx} + y(1) = 0$$x \frac{dy}{dx} = -y$$\frac{dy}{dx} = -\frac{y}{x}$

If $xy = \tan^{-1}(xy) + \cot^{-1}(xy)$,then $\frac{dy}{dx}$ is equal to

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