Question

derivative of y = ( x +sinx ) / ( x + cos x )
$y = \frac{( x +sinx)}{( x + cos x )}$
w.r.t x ​

Answer 1

$\green{ \large \sf \: Given :-}$

$\sf \: y = \frac{x + \sin(x) }{x + \cos(x) }$

$\green{ \large \sf \: Solution :-}$

$: \implies\sf \: y = \frac{x + \sin(x) }{x + \cos(x) }$

$: \implies\sf \: \frac{dy}{dx} = \frac{d}{dx} \bigg(\frac{x + \sin(x) }{x + \cos(x) } \bigg)$

$: \implies\sf \: \frac{dy}{dx} = \frac{ \frac{d}{dx} (x + \sin x)(x + \cos x) - \frac{d}{dx} (x + \cos x)(x + \sin x) }{(x + \cos x ) {}^{2} }$

$: \implies\sf \: \frac{dy}{dx} = \frac{ (1 + \cos x)(x + \cos x) - (1 - \sin x)(x + \sin x) }{(x + \cos x ) {}^{2} }$

$: \implies\sf \: \frac{dy}{dx} = \frac{ x + \cos x + x \cos x + { \cos ^{2}x } - (x + \sin x - x \sin x -\cos ^{2}x ) }{(x + \cos x ) {}^{2} }$

$: \implies\sf \: \frac{dy}{dx} = \frac{ x + \cos x + x \cos x + { \cos ^{2}x } - x - \sin x + x \sin x + \sin ^{2}x ) }{(x + \cos x ) {}^{2} }$

$: \implies\sf \: \frac{dy}{dx} = \frac{ \cos x + x \cos x + { \cos ^{2}x } +\sin ^{2}x - \sin x + x \sin x ) }{(x + \cos x ) {}^{2} }$

$\sf \: Formula \: \: { \cos ^{2}x } +\sin ^{2}x = 1$

$: \implies\sf \: \frac{dy}{dx} = \frac{ \cos x + x \cos x - \sin x + x \sin x ) }{(x + \cos x ) {}^{2} }$

Hope it help.