Question

Find dy/dx given that y = sin x / 1 + cos x​

Answer 1

Answer:

$\frac{1}{1 + \cos(x) }$

Step-by-step explanation:

$y = \frac{ \sin(x) }{1 + \cos(x) }$

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"

$(\frac{u}{v} )' = \frac{v. \: (u)' - u.(v)' }{ {v}^{2} }$

here,

• since u is the numerator u= sinx and u = cos x
• v(denominator) = 1 + cos x; v' = - sinx

$= \frac{((1 + \cos \: x) \cos \: x )- (\sin \: x. ( - \sin \: x) ) }{( {1 + \cos(x)) }^{2} }$

$= \frac{ \cos(x) + \cos {}^{2} (x) + \sin {}^{2} (x) }{(1 + \cos \: x) {}^{2} }$

since cos²x + sin²x = 1

$= \frac{ \cos \: x + 1}{(1 + \cos \: x) {}^{2} }$

diving numerator and denominator by 1 + cos x

$= \frac{1}{1 + \cos(x) }$