What is the derivative of # y=sinx/(2+cosx)#?

Question

Cameron Alexander · Accepted Answer

#y' = frac(1 + 2 cos(x))((2 + cos(x))^(2))#

We have: #y = frac(sin(x))(2 + cos(x))#

This function can be differentiated using the "quotient rule":

#Rightarrow y' = frac((2 + cos(x)) cdot frac(d)(dx)(sin(x)) - (sin(x)) cdot frac(d)(dx)(2 + cos(x)))((2 + cos(x))^(2))#

#Rightarrow y' = frac((2 + cos(x)) cdot cos(x) - sin(x) cdot (- sin(x)))((2 + cos(x))^(2))#

#Rightarrow y' = frac(2 cos(x) + cos^(2)(x) + sin^(2)(x))((2 + cos(x))^(2))#

One of the Pythagorean identities is #cos^(2)(x) + sin^(2)(x) = 1#.

We can rearrange it to get:

#Rightarrow sin^(2)(x) = 1 - cos^(2)(x)#

Let's apply this rearranged identity:

#Rightarrow y' = frac(2 cos(x) + cos^(2)(x) + 1 - cos^(2)(x))((2 + cos(x))^(2))#

#Rightarrow y' = frac(1 + 2 cos(x))((2 + cos(x))^(2))#

Adam Adkins · Answer

undefined To find the derivative of $ y = \frac{\sin(x)}{2 + \cos(x)} $, you can use the quotient rule, which states that if $ y = \frac{u}{v} $, then $ y' = \frac{u'v - uv'}{v^2} $, where $ u' $ and $ v' $ are the derivatives of $ u $ and $ v $ respectively.

Let $ u = \sin(x) $ and $ v = 2 + \cos(x) $.

Then, $ u' = \cos(x) $ and $ v' = -\sin(x) $.

Now apply the quotient rule:

$$ y' = \frac{\sin(x)(-\sin(x)) - \cos(x)\cos(x)}{(2 + \cos(x))^2} $$

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

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

So, the derivative of $ y = \frac{\sin(x)}{2 + \cos(x)} $ is $ y' = \frac{-1}{(2 + \cos(x))^2} $.