Question

Find differentiation of x-y = tanxy​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: x - y = tan(xy)$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}(x - y) = \dfrac{d}{dx}tan(xy)$

$\rm \: \dfrac{d}{dx}x - \dfrac{d}{dx}y = {sec}^{2}(xy)\dfrac{d}{dx}(xy)$

$\rm \: 1 - \dfrac{dy}{dx} = {sec}^{2}(xy)\bigg(x\dfrac{d}{dx}y + y\dfrac{d}{dx}x\bigg)$

$\rm \: 1 - \dfrac{dy}{dx} = {sec}^{2}(xy)\bigg(x\dfrac{dy}{dx} + y \times 1\bigg)$

$\rm \: 1 - \dfrac{dy}{dx} = {sec}^{2}(xy)\bigg(x\dfrac{dy}{dx} + y \bigg)$

$\rm \: 1 - \dfrac{dy}{dx} = x{sec}^{2}(xy) \dfrac{dy}{dx} + y {sec}^{2}(xy)$

$\rm \: \dfrac{dy}{dx} + x{sec}^{2}(xy) \dfrac{dy}{dx} = 1 - y {sec}^{2}(xy)$

$\rm \: \bigg(1+ x{sec}^{2}(xy) \bigg)\dfrac{dy}{dx} = 1 - y {sec}^{2}(xy)$

$\rm\implies \:\rm \: \dfrac{dy}{dx} = \dfrac{1 - y {sec}^{2}(xy)}{1+ x{sec}^{2}(xy)}$

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Formulae Used :-

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx} tanx = {sec}^{2} x \: \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx}(u.v) = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$