Question

Class _12 Differentiation of implicit functions.

If xy= tan ( xy )
show that ,

dy/dx = -y/x
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Answer 1

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$xy = \tan(xy)$

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$xy = \tan(xy) \\ \implies y \times 1 + \frac{dy}{dx} \times x = { \sec }^{2} (xy){x \times \frac{dy}{dx} + y \times 1} \\ \implies y + x \frac{dy}{dx} = y {sec}^{2} (xy) + x \frac{dy}{dx} { \sec}^{2} (xy) \\ \implies x \frac{dy}{dx} - x \frac{dy}{dx} { \sec }^{2} (xy) = y { \sec }^{2} (xy) - y \\ \implies x \frac{dy}{dx} (1 - { \sec }^{2} (xy)) = y( { \sec}^{2} (xy) - 1) \\ \implies x \frac{dy}{dx} = - y \times \frac{1 - { \sec }^{2} (xy) }{1 - { \sec}^{2} (xy)} \\ \implies \frac{dy}{dx} = - \frac{y}{x}$

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Answer 2

Answer:

Please see the attachment