Answer:
[tex] \frac{1}{1 + \cos(x) } [/tex]
Step-by-step explanation:
[tex]y = \frac{ \sin(x) }{1 + \cos(x) } [/tex]
differentiating numerator wrt x :-
(sinx)' = cos x
differentiating denominator wrt x :-
(1 + cos x)' = (cosx)' = - sinx
- Let's say the denominator was "v" and the numerator was "u"
[tex] (\frac{u}{v} )' = \frac{v. \: (u)' - u.(v)' }{ {v}^{2} } [/tex]
here,
- since u is the numerator u= sinx and u = cos x
- v(denominator) = 1 + cos x; v' = - sinx
[tex] = \frac{((1 + \cos \: x) \cos \: x )- (\sin \: x. ( - \sin \: x) ) }{( {1 + \cos(x)) }^{2} } [/tex]
[tex] = \frac{ \cos(x) + \cos {}^{2} (x) + \sin {}^{2} (x) }{(1 + \cos \: x) {}^{2} } [/tex]
since cos²x + sin²x = 1
[tex] = \frac{ \cos \: x + 1}{(1 + \cos \: x) {}^{2} } [/tex]
diving numerator and denominator by 1 + cos x
[tex] = \frac{1}{1 + \cos(x) } [/tex]
