Question

$\large{ \underline{ \underline{ \bigstar \: \: \: \: \pmb{ \sf{Question \: : }}}}}$
$\odot \: \: \: \: ɪғ \: \: \sf{ {x}^{5} + {y}^{5} = {x}^{3} {y}^{2} }, sʜᴏᴡ \: \: ᴛʜᴀᴛ \: \: \sf { \frac{dy}{dx} = \frac{y}{x} }$
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Answer 1

Appropriate Question :-

$\rm :\longmapsto\:ɪғ \: {(x + y)}^{5} = {x}^{3} {y}^{2}, \: sʜᴏᴡ \: ᴛʜᴀᴛ \: \dfrac{dy}{dx} = \dfrac{y}{x}$

Formula Used :-

$\red{ \boxed{ \sf{ \: log(xy) = logx + logy }}}$

$\red{ \boxed{ \sf{ log( {x}^{y} ) \: = \: y \: logx}}}$

$\red{ \boxed{ \sf{ \dfrac{d}{dx}x\: = \: 1}}}$

$\red{ \boxed{ \sf{ \dfrac{d}{dx}logx\: = \: \frac{1}{x} }}}$

Solution :-

Given function is

$\rm :\longmapsto\: {(x + y)}^{5} = {x}^{3} {y}^{2}$

On taking log both sides, we get

$\rm :\longmapsto\:log {(x + y)}^{5} =log( {x}^{3} {y}^{2} )$

$\rm :\longmapsto\:5 log(x + y) = log( {x}^{3} ) + log( {y}^{2} )$

$\rm :\longmapsto\:5 log(x + y) = 3 log( {x}) + 2log( {y} )$

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

$\rm :\longmapsto\:\dfrac{d}{dx}5 log(x + y) = \dfrac{d}{dx}3 log( {x}) + \dfrac{d}{dx} 2log( {y} )$

$\rm :\longmapsto\:5\dfrac{d}{dx} log(x + y) = 3\dfrac{d}{dx}log( {x}) +2 \dfrac{d}{dx} log( {y} )$

$\rm :\longmapsto\:5 \times \dfrac{1}{x + y} \bigg(1 + \dfrac{dy}{dx} \bigg) = \dfrac{3}{x} + \dfrac{2}{y}\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{5}{x + y} \bigg(1 + \dfrac{dy}{dx} \bigg) = \dfrac{3}{x} + \dfrac{2}{y}\dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{5}{x + y} + \dfrac{5}{x + y} \dfrac{dy}{dx} = \dfrac{3}{x} + \dfrac{2}{y}\dfrac{dy}{dx}$

$\rm :\longmapsto\: \dfrac{5}{x + y} \dfrac{dy}{dx} - \dfrac{2}{y}\dfrac{dy}{dx}= \dfrac{3}{x} - \dfrac{5}{x + y}$

$\rm :\longmapsto\: \bigg(\dfrac{5}{x + y} - \dfrac{2}{y} \bigg)\dfrac{dy}{dx}= \dfrac{3}{x} - \dfrac{5}{x + y}$

$\rm :\longmapsto\: \bigg(\dfrac{5y - 2x - 2y}{y(x + y)} \bigg)\dfrac{dy}{dx}= \dfrac{3x + 3y - 5x}{x(x + y)}$

$\rm :\longmapsto\: \bigg(\dfrac{3y - 2x}{y(x + y)} \bigg)\dfrac{dy}{dx}= \dfrac{3y - 2x}{x(x + y)}$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{x}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{y}{x}$

Short Cut Trick

$\boxed{\bf :\longmapsto\:ɪғ \: {(x + y)}^{m + n} = {x}^{m} {y}^{n}, \: then\: \dfrac{dy}{dx} = \dfrac{y}{x} }$