Question
Giving your answer in a further factored format differentiate xsinx/1+cosx with respect to x
Answers
GPT-4o mini
To differentiate the function \( \frac{x \sin x}{1 + \cos x} \) with respect to \( x \), we can use the quotient rule. The quotient rule states that if we have a function of the form \( \frac{u}{v} \), where \( u \) and \( v \) are both functions of \( x \), then the derivative is given by:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
In this case, we have:
- \( u = x \sin x \)
- \( v = 1 + \cos x \)
### Step 1: Differentiate \( u \) and \( v \)
First, we need to compute \( u' \) and \( v' \):
1. **Differentiating \( u = x \sin x \)**:
\[
u' = \sin x + x \cos x \quad \text{(using the product rule)}
\]
2. **Differentiating \( v = 1 + \cos x \)**:
\[
v' = -\sin x
\]
### Step 2: Applying the Quotient Rule
Now we can apply the quotient rule:
\[
\frac{d}{dx} \left( \frac{x \sin x}{1 + \cos x} \right) = \frac{(\sin x + x \cos x)(1 + \cos x) - (x \sin x)(-\sin x)}{(1 + \cos x)^2}
\]
### Step 3: Simplifying the Numerator
Let's simplify the numerator:
\[
(\sin x + x \cos x)(1 + \cos x) + x \sin^2 x
\]
Expanding it:
\[
= \sin x + \sin x \cos x + x \cos x + x \cos^2 x + x \sin^2 x
\]
### Step 4: Factor the Numerator
Now we can group the terms to factor:
The numerator becomes:
\[
\sin x + x \cos x + x \cos^2 x + x \sin^2 x + \sin x \cos x
\]
Notice that \( x (\cos^2 x + \sin^2 x) = x \) (since \( \cos^2 x + \sin^2 x = 1 \)), thus we can write:
\[
= \sin x + x + x \cos x + \sin x \cos x
\]
Now we can factor out \( \sin x \) from the first two terms and rearrange terms:
\[
= \sin x (1 + \cos x) + x (1 + \cos x)
\]
Factoring out \( (1 + \cos x) \):
\[
= (1 + \cos x)(\sin x + x)
\]
### Final Result
Putting everything back into the equation gives us:
\[
\frac{d}{dx} \left( \frac{x \sin x}{1 + \cos x} \right) = \frac{(1 + \cos x)(\sin x + x)}{(1 + \cos x)^2}
\]
We can simplify this to get:
\[
\frac{\sin x + x}{1 + \cos x}
\]
Thus, the final answer is:
\[
\frac{d}{dx} \left( \frac{x \sin x}{1 + \cos x} \right) = \frac{\sin x + x}{1 + \cos x}
\]
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
In this case, we have:
- \( u = x \sin x \)
- \( v = 1 + \cos x \)
### Step 1: Differentiate \( u \) and \( v \)
First, we need to compute \( u' \) and \( v' \):
1. **Differentiating \( u = x \sin x \)**:
\[
u' = \sin x + x \cos x \quad \text{(using the product rule)}
\]
2. **Differentiating \( v = 1 + \cos x \)**:
\[
v' = -\sin x
\]
### Step 2: Applying the Quotient Rule
Now we can apply the quotient rule:
\[
\frac{d}{dx} \left( \frac{x \sin x}{1 + \cos x} \right) = \frac{(\sin x + x \cos x)(1 + \cos x) - (x \sin x)(-\sin x)}{(1 + \cos x)^2}
\]
### Step 3: Simplifying the Numerator
Let's simplify the numerator:
\[
(\sin x + x \cos x)(1 + \cos x) + x \sin^2 x
\]
Expanding it:
\[
= \sin x + \sin x \cos x + x \cos x + x \cos^2 x + x \sin^2 x
\]
### Step 4: Factor the Numerator
Now we can group the terms to factor:
The numerator becomes:
\[
\sin x + x \cos x + x \cos^2 x + x \sin^2 x + \sin x \cos x
\]
Notice that \( x (\cos^2 x + \sin^2 x) = x \) (since \( \cos^2 x + \sin^2 x = 1 \)), thus we can write:
\[
= \sin x + x + x \cos x + \sin x \cos x
\]
Now we can factor out \( \sin x \) from the first two terms and rearrange terms:
\[
= \sin x (1 + \cos x) + x (1 + \cos x)
\]
Factoring out \( (1 + \cos x) \):
\[
= (1 + \cos x)(\sin x + x)
\]
### Final Result
Putting everything back into the equation gives us:
\[
\frac{d}{dx} \left( \frac{x \sin x}{1 + \cos x} \right) = \frac{(1 + \cos x)(\sin x + x)}{(1 + \cos x)^2}
\]
We can simplify this to get:
\[
\frac{\sin x + x}{1 + \cos x}
\]
Thus, the final answer is:
\[
\frac{d}{dx} \left( \frac{x \sin x}{1 + \cos x} \right) = \frac{\sin x + x}{1 + \cos x}
\]