Evaluate the derivative of the following function at indicated point :
`(1-sinx)/(1+cosx), x=pi/2`.

Answer

To evaluate the derivative of the function \( f(x) = \frac{1 - \sin x}{1 + \cos x} \) at the point \( x = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Differentiate the function using the Quotient Rule
The Quotient Rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

For our function:
- \( u = 1 - \sin x \)
- \( v = 1 + \cos x \)

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

### Step 2: Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
- \( \frac{du}{dx} = \frac{d}{dx}(1 - \sin x) = 0 - \cos x = -\cos x \)
- \( \frac{dv}{dx} = \frac{d}{dx}(1 + \cos x) = 0 - \sin x = -\sin x \)

### Step 3: Substitute into the Quotient Rule formula
Now substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the Quotient Rule formula:

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

### Step 4: Simplify the expression
Now, simplify the numerator:

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

Expanding the numerator:

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

Using the identity \( \sin^2 x + \cos^2 x = 1 \):

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

This simplifies to:

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

### Step 5: Evaluate at \( x = \frac{\pi}{2} \)
Now substitute \( x = \frac{\pi}{2} \) into the derivative:

\[ \frac{dy}{dx}\bigg|_{x=\frac{\pi}{2}} = \frac{\sin\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right) - 1}{(1 + \cos\left(\frac{\pi}{2}\right))^2} \]

Calculating the values:

- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Substituting these values:

\[ = \frac{1 - 0 - 1}{(1 + 0)^2} = \frac{0}{1} = 0 \]

### Final Answer
Thus, the derivative of the function at \( x = \frac{\pi}{2} \) is:

\[ \frac{dy}{dx}\bigg|_{x=\frac{\pi}{2}} = 0 \]