If siny = xsin(x+y)Then, prove dy/dx= {sin(x+y)+xcos(x+y)} / {cosy-xco

1.	If siny = xsin(x+y)Then, prove dy/dx= {sin(x+y)+xcos(x+y)} / {cosy-xcos(x+y)}
Answer» $\large\underline{\sf{Given \:Question - }}$ $\sf \: siny = <klux>X</klux> \: sin(x + y)$ $\large\underline{\sf{To\:Prove - }}$ $\sf \: \dfrac{dy}{dx} = \dfrac{sin(x + y) + x \: cos(x + y)}{cosy - xcos(x + y)}$ $\large\begin{gathered}{\sf{{\underline{Formula \: Used - }}}} \end{gathered}$ $1. \: \: \boxed{ \bf{ \: \dfrac{d}{dx}u.v = v\dfrac{d}{dx}u + u\dfrac{d}{dx}v}}$ $2. \: \: \boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}$ $3 \: \: \boxed{ \bf{ \: \dfrac{d}{dx}cosx = - \: sinx}}$ $4. \: \: \boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}$ $\large\underline{\sf{Solution-}}$ Given that $\rm :\longmapsto\: \bf \: siny = x \: sin(x + y)$ Differentiating both SIDES W. r. t. x, we get $\rm :\longmapsto\:\dfrac{d}{dx}siny = \dfrac{d}{dx}\bigg( x \: sin(x + y)\bigg)$ $\rm :\longmapsto\:cosy\dfrac{dy}{dx} = x\dfrac{d}{dx}sin(x + y) + sin(x + y)\dfrac{d}{dx}x$ $\rm :\longmapsto\:cosy\dfrac{dy}{dx} = xcos(x + y)\dfrac{d}{dx}(x + y) + sin(x + y)$ $\rm :\longmapsto\:cosy\dfrac{dy}{dx} = xcos(x + y)\bigg(1 + \dfrac{dy}{dx} \bigg) + sin(x + y)$ $\rm :\longmapsto\:cosy\dfrac{dy}{dx} = xcos(x + y) + xcos(x + y)\dfrac{dy}{dx} + sin(x + y)$ $\rm :\longmapsto\:\dfrac{dy}{dx}\bigg(cosy - xcos(x + y) \bigg) = xcos(x + y) + sin(x + y)$ $\bf\implies \: \: \dfrac{dy}{dx} = \dfrac{sin(x + y) + x \: cos(x + y)}{cosy - xcos(x + y)}$ ${\boxed{\bf{Hence, Proved}}}$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional INFORMATION :- $\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{ n- 1} }}$ $\boxed{ \bf{ \: \dfrac{d}{dx}logx = \dfrac{1}{x} }}$ $\boxed{ \bf{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x} }}$ $\boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} loga}}$ $\boxed{ \bf{ \: tanx = {sec}^{2} x}}$ $\boxed{ \bf{ \: \dfrac{d}{dx}cotx = - {cosec}^{2} x}}$ $\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}$