Question

find dy/dx, if e^siny=xy​

Answer 1

Answer:

Your answer attached in the photo

Answer 2

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

$\rm :\longmapsto\: {e}^{siny} = xy$

Taking log on both sides, we get

$\rm :\longmapsto\: log \: ({e}^{siny}) = log(xy)$

$\rm :\longmapsto\:siny \: loge \: = \: logx \: + \: logy$

$\red{\bigg \{ \because \:log {x}^{y} = ylogx \: \bigg \}} \\ \red{\bigg \{ \because \: \: log(xy) = logx + logy\bigg \}}$

$\rm :\longmapsto\:\:siny \: = \: logx \: + \: logy$

$\red{\bigg \{ \because \: loge = 1\bigg \}}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} \:siny \: = \dfrac{d}{dx}( \: logx \: + \: logy)$

$\rm :\longmapsto\:cosy\dfrac{dy}{dx} \: = \dfrac{d}{dx}\: logx \: + \dfrac{d}{dx}\: logy$

$\green{\boxed{ \because \: \bf \:\dfrac{d}{dx}sinx = cosx \: }}$

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

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

$\rm :\longmapsto\: \bigg(cosy \: - \: \dfrac{1}{y} \bigg)\dfrac{dy}{dx} = \dfrac{1}{x}$

$\rm :\longmapsto\: \bigg(\: \dfrac{y \: cosy - 1}{y} \bigg)\dfrac{dy}{dx} = \dfrac{1}{x}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{y}{x(y \: cosy \: - \: 1)}$

Additional Information :-

$\green{\boxed{ \: \bf \:\dfrac{d}{dx}cosx = - \: sinx \: }}$

$\green{\boxed{ \: \bf \:\dfrac{d}{dx}cotx = - \: {cosec}^{2} x \: }}$

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

$\green{\boxed{ \bf \: \dfrac{d}{dx}secx = secx \: tanx\: }}$

$\green{\boxed{ \bf \: \dfrac{d}{dx}cosecx = - \: cosecx \: cotx\: }}$

$\green{\boxed{ \bf \:\dfrac{d}{dx}k = 0 \: }}$

$\green{\boxed{ \bf \:\dfrac{d}{dx}x= 1 \: }}$

$\green{\boxed{ \bf \:\dfrac{d}{dx}logx= \frac{1}{x} \: }}$

$\green{\boxed{ \bf \: \dfrac{d}{dx} {e}^{x} \: = \: {e}^{x} }}$

$\green{\boxed{ \bf \: \dfrac{d}{dx} {a}^{x} \: = \: {a}^{x} log(a) }}$