Question
Giving your answer in advanced format differentiate xsinx/1+cosx
Answers
GPT-4o mini
To differentiate the function
\[
y = \frac{x \sin x}{1 + \cos x}
\]
we will use the quotient rule, which states that if you have a function in the form \(y = \frac{u}{v}\), then the derivative \(y'\) is given by:
\[
y' = \frac{u'v - uv'}{v^2}
\]
where \(u = x \sin x\) and \(v = 1 + \cos x\).
### Step 1: Differentiate \(u\) and \(v\)
1. First, we need to differentiate \(u = x \sin x\):
Using the product rule:
\[
u' = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x)
\]
\[
u' = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x
\]
2. Now, differentiate \(v = 1 + \cos x\):
\[
v' = \frac{d}{dx}(1) + \frac{d}{dx}(\cos x) = 0 - \sin x = -\sin x
\]
### Step 2: Apply the Quotient Rule
Now we apply the quotient rule:
\[
y' = \frac{u'v - uv'}{v^2}
\]
Substituting \(u\), \(u'\), \(v\), and \(v'\) into the formula:
\[
y' = \frac{(\sin x + x \cos x)(1 + \cos x) - (x \sin x)(-\sin x)}{(1 + \cos x)^2}
\]
### Step 3: Simplify the Expression
Now we simplify the numerator:
1. Expand \(u'v\):
\[
(\sin x + x \cos x)(1 + \cos x) = \sin x + \sin x \cos x + x \cos x + x \cos^2 x
\]
2. Simplify \(uv'\):
\[
-uv' = - (x \sin x)(-\sin x) = x \sin^2 x
\]
Now we can combine the expressions:
\[
y' = \frac{\sin x + \sin x \cos x + x \cos x + x \cos^2 x + x \sin^2 x}{(1 + \cos x)^2}
\]
### Final Result
Thus, the derivative of the function \(y = \frac{x \sin x}{1 + \cos x}\) is:
\[
y' = \frac{\sin x (1 + \cos x) + x \cos x + x \cos^2 x + x \sin^2 x}{(1 + \cos x)^2}
\]
You can leave the answer in this form, or you can further factor or simplify the numerator if desired, but this is the differentiated form using the quotient rule.
\[
y = \frac{x \sin x}{1 + \cos x}
\]
we will use the quotient rule, which states that if you have a function in the form \(y = \frac{u}{v}\), then the derivative \(y'\) is given by:
\[
y' = \frac{u'v - uv'}{v^2}
\]
where \(u = x \sin x\) and \(v = 1 + \cos x\).
### Step 1: Differentiate \(u\) and \(v\)
1. First, we need to differentiate \(u = x \sin x\):
Using the product rule:
\[
u' = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x)
\]
\[
u' = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x
\]
2. Now, differentiate \(v = 1 + \cos x\):
\[
v' = \frac{d}{dx}(1) + \frac{d}{dx}(\cos x) = 0 - \sin x = -\sin x
\]
### Step 2: Apply the Quotient Rule
Now we apply the quotient rule:
\[
y' = \frac{u'v - uv'}{v^2}
\]
Substituting \(u\), \(u'\), \(v\), and \(v'\) into the formula:
\[
y' = \frac{(\sin x + x \cos x)(1 + \cos x) - (x \sin x)(-\sin x)}{(1 + \cos x)^2}
\]
### Step 3: Simplify the Expression
Now we simplify the numerator:
1. Expand \(u'v\):
\[
(\sin x + x \cos x)(1 + \cos x) = \sin x + \sin x \cos x + x \cos x + x \cos^2 x
\]
2. Simplify \(uv'\):
\[
-uv' = - (x \sin x)(-\sin x) = x \sin^2 x
\]
Now we can combine the expressions:
\[
y' = \frac{\sin x + \sin x \cos x + x \cos x + x \cos^2 x + x \sin^2 x}{(1 + \cos x)^2}
\]
### Final Result
Thus, the derivative of the function \(y = \frac{x \sin x}{1 + \cos x}\) is:
\[
y' = \frac{\sin x (1 + \cos x) + x \cos x + x \cos^2 x + x \sin^2 x}{(1 + \cos x)^2}
\]
You can leave the answer in this form, or you can further factor or simplify the numerator if desired, but this is the differentiated form using the quotient rule.