Answer:
![(1)/(1 + \cos(x) )](https://img.qammunity.org/2022/formulas/mathematics/college/jqazur21qemg0uuz64rrvxalls3slv65gh.png)
Explanation:
![y = ( \sin(x) )/(1 + \cos(x) )](https://img.qammunity.org/2022/formulas/mathematics/college/zxdahq8w4piudpcfu694salx6ilxvyklzr.png)
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"
![((u)/(v) )' = \frac{v. \: (u)' - u.(v)' }{ {v}^(2) }](https://img.qammunity.org/2022/formulas/mathematics/college/kembe4efokls6hs8q5bd2giy2jngghwdqu.png)
here,
- since u is the numerator u= sinx and u = cos x
- v(denominator) = 1 + cos x; v' = - sinx
![= \frac{((1 + \cos \: x) \cos \: x )- (\sin \: x. ( - \sin \: x) ) }{( {1 + \cos(x)) }^(2) }](https://img.qammunity.org/2022/formulas/mathematics/college/ryvzb0mr8zery68r05eagr60yeo9astrxn.png)
![= \frac{ \cos(x) + \cos {}^(2) (x) + \sin {}^(2) (x) }{(1 + \cos \: x) {}^(2) }](https://img.qammunity.org/2022/formulas/mathematics/college/jt3ff8zhvikb98nrokk0dnbwqri5hxkbw5.png)
since cos²x + sin²x = 1
![= \frac{ \cos \: x + 1}{(1 + \cos \: x) {}^(2) }](https://img.qammunity.org/2022/formulas/mathematics/college/5scoquarm837ye9nqea8df3qrq5krqsqdw.png)
diving numerator and denominator by 1 + cos x
![= (1)/(1 + \cos(x) )](https://img.qammunity.org/2022/formulas/mathematics/college/v1e3bbyfoyc3xa8ssjty5k21tg7mjnwwr1.png)