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

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Question

(A) We know the trigonometric identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.Given the equation $xy = 	an^{-1}(xy)

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$\frac{-1}{2}$

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(A) We know the trigonometric identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.Given the equation $xy = 	an^{-1}(xy) + \cot^{-1}(xy)$,we can substitute the identity to get $xy = \frac{\pi}{2}$.Differentiating both sides with respect to $x$ using the product rule:$y + x \frac{dy}{dx} = 0$.Solving for $\frac{dy}{dx}$,we get $\frac{dy}{dx} = -\frac{y}{x}$.Evaluating this at the point $(4, 2)$:$\left(\frac{dy}{dx}ight)_{(4,2)} = -\frac{2}{4} = -\frac{1}{2}$.

If $xy = \tan^{-1}(xy) + \cot^{-1}(xy)$,then $\left(\frac{dy}{dx}\right)_{(4,2)} = ?$ (where $x, y \in \mathbb{R}$)

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