Question

find dy/dx if x^2y^2=tan(xy)
​

Answer 1

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

Given function is

$\rm :\longmapsto\: {x}^{2} {y}^{2} = tan(xy)$

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

$\rm :\longmapsto\: \dfrac{d}{dx}[ \: {x}^{2} {y}^{2} \: ] =\dfrac{d}{dx} tan(xy)$

We know,

$\green{ \boxed{ \sf{ \:\dfrac{d}{dx}uv = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}}}$

and

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

So, using these, we get

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

We know,

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

So, using this, we get

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

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

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

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

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{{sec}^{2}(xy) - 2 {xy}^{2} }{ {2yx}^{2} - x{sec}^{2}(xy)}$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \: \: \: \: \: \:\dfrac{dy}{dx} = \dfrac{{sec}^{2}(xy) - 2 {xy}^{2} }{ {2yx}^{2} - x{sec}^{2}(xy)} \: \: \: \: \: }}}$

Additional Information :-

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