Find the differential coefficient of the following functions: `(cos x)/(1+cos x)`

Answer

To find the differential coefficient of the function \( f(x) = \frac{\cos x}{1 + \cos x} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), then the derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step 1: Identify \( u \) and \( v \)
In our case:
- \( u = \cos x \)
- \( v = 1 + \cos x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find the derivatives of \( u \) and \( v \):
- \( \frac{du}{dx} = -\sin x \)
- \( \frac{dv}{dx} = 0 - \sin x = -\sin x \)

### Step 3: Apply the Quotient Rule
Now we apply the quotient rule:

\[ \frac{d}{dx}\left(\frac{\cos x}{1 + \cos x}\right) = \frac{(1 + \cos x)(-\sin x) - (\cos x)(-\sin x)}{(1 + \cos x)^2} \]

### Step 4: Simplify the Expression
Now we simplify the numerator:

\[ = \frac{-(1 + \cos x)\sin x + \cos x \sin x}{(1 + \cos x)^2} \]

Distributing the terms in the numerator:

\[ = \frac{-\sin x - \cos x \sin x + \cos x \sin x}{(1 + \cos x)^2} \]

The \( -\cos x \sin x \) and \( +\cos x \sin x \) cancel each other out:

\[ = \frac{-\sin x}{(1 + \cos x)^2} \]

### Final Result
Thus, the differential coefficient of the function \( f(x) = \frac{\cos x}{1 + \cos x} \) is:

\[ f'(x) = \frac{-\sin x}{(1 + \cos x)^2} \]