Question
Giving your answer in a further factored format differentiate xsinx/1+cosx
Answers
GPT-4o mini
To differentiate the function \( y = \frac{x \sin x}{1 + \cos x} \), we can apply the quotient rule. The quotient rule states that if you have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative \( y' \) is given by:
\[
y' = \frac{u'v - uv'}{v^2}
\]
In our case:
- \( u = x \sin x \)
- \( v = 1 + \cos x \)
Now we need to find \( u' \) and \( v' \):
1. **Differentiate \( u \):**
Using the product rule:
\[
u' = \sin x + x \cos x
\]
2. **Differentiate \( v \):**
\[
v' = -\sin x
\]
Now we have:
- \( u' = \sin x + x \cos x \)
- \( v' = -\sin x \)
Now we can apply the quotient rule:
\[
y' = \frac{(\sin x + x \cos x)(1 + \cos x) - (x \sin x)(-\sin x)}{(1 + \cos x)^2}
\]
Now simplify the numerator:
\[
= (\sin x + x \cos x)(1 + \cos x) + x \sin^2 x
\]
Expanding the first term:
\[
= \sin x + \sin x \cos x + x \cos x + x \cos^2 x + x \sin^2 x
\]
Thus, combining like terms gives:
\[
= \sin x + x \cos x + x \cos^2 x + \sin x \cos x + x \sin^2 x
\]
So, our first derivative can be expressed as:
\[
y' = \frac{\sin x + x \cos x + x \cos^2 x + \sin x \cos x + x \sin^2 x}{(1 + \cos x)^2}
\]
Now, we can further factor some parts of the numerator, noting that \( x \cos^2 x + x \sin^2 x = x \) since \( \cos^2 x + \sin^2 x = 1 \):
So we have:
\[
y' = \frac{\sin x + x(1) + x \cos x \sin x}{(1 + \cos x)^2}
\]
Final factored form, simplifying where possible, leads to:
\[
y' = \frac{\sin x + x + x \cos x \sin x}{(1 + \cos x)^2}
\]
This is the derivative of the given function \( y = \frac{x \sin x}{1 + \cos x} \). Further simplification can depend on context or additional instructions but this expression captures the critical components of the differentiation process.
\[
y' = \frac{u'v - uv'}{v^2}
\]
In our case:
- \( u = x \sin x \)
- \( v = 1 + \cos x \)
Now we need to find \( u' \) and \( v' \):
1. **Differentiate \( u \):**
Using the product rule:
\[
u' = \sin x + x \cos x
\]
2. **Differentiate \( v \):**
\[
v' = -\sin x
\]
Now we have:
- \( u' = \sin x + x \cos x \)
- \( v' = -\sin x \)
Now we can apply the quotient rule:
\[
y' = \frac{(\sin x + x \cos x)(1 + \cos x) - (x \sin x)(-\sin x)}{(1 + \cos x)^2}
\]
Now simplify the numerator:
\[
= (\sin x + x \cos x)(1 + \cos x) + x \sin^2 x
\]
Expanding the first term:
\[
= \sin x + \sin x \cos x + x \cos x + x \cos^2 x + x \sin^2 x
\]
Thus, combining like terms gives:
\[
= \sin x + x \cos x + x \cos^2 x + \sin x \cos x + x \sin^2 x
\]
So, our first derivative can be expressed as:
\[
y' = \frac{\sin x + x \cos x + x \cos^2 x + \sin x \cos x + x \sin^2 x}{(1 + \cos x)^2}
\]
Now, we can further factor some parts of the numerator, noting that \( x \cos^2 x + x \sin^2 x = x \) since \( \cos^2 x + \sin^2 x = 1 \):
So we have:
\[
y' = \frac{\sin x + x(1) + x \cos x \sin x}{(1 + \cos x)^2}
\]
Final factored form, simplifying where possible, leads to:
\[
y' = \frac{\sin x + x + x \cos x \sin x}{(1 + \cos x)^2}
\]
This is the derivative of the given function \( y = \frac{x \sin x}{1 + \cos x} \). Further simplification can depend on context or additional instructions but this expression captures the critical components of the differentiation process.