Differentiate `cot ^(-1) ((cos x )/(1 +sin x )) ` w.r. to x

Answer

To differentiate the function \( y = \cot^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the Function
Let
\[ y = \cot^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \]

### Step 2: Use the Cotangent Identity
Recall that \( \cot^{-1}(u) \) can be rewritten using the identity:
\[ \cot^{-1}(u) = \tan^{-1}\left(\frac{1}{u}\right) \]
Thus, we can write:
\[ y = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right) \]

### Step 3: Differentiate using the Chain Rule
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{1 + \sin x}{\cos x}\right)^2} \cdot \frac{d}{dx}\left(\frac{1 + \sin x}{\cos x}\right) \]

### Step 4: Differentiate the Inner Function
Let \( u = \frac{1 + \sin x}{\cos x} \). We need to differentiate \( u \):
Using the quotient rule:
\[ \frac{du}{dx} = \frac{\cos x \cdot \frac{d}{dx}(1 + \sin x) - (1 + \sin x) \cdot \frac{d}{dx}(\cos x)}{\cos^2 x} \]
Calculating the derivatives:
- \( \frac{d}{dx}(1 + \sin x) = \cos x \)
- \( \frac{d}{dx}(\cos x) = -\sin x \)

Substituting these into the quotient rule gives:
\[ \frac{du}{dx} = \frac{\cos x \cdot \cos x - (1 + \sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + (1 + \sin x)\sin x}{\cos^2 x} \]

### Step 5: Substitute back into the Derivative
Now we substitute \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{1 + \sin x}{\cos x}\right)^2} \cdot \frac{\cos^2 x + (1 + \sin x)\sin x}{\cos^2 x} \]

### Step 6: Simplify the Expression
To simplify \( \frac{dy}{dx} \), we can use the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \):
\[ 1 + \left(\frac{1 + \sin x}{\cos x}\right)^2 = \frac{\cos^2 x + (1 + \sin x)^2}{\cos^2 x} \]
Thus, we can simplify:
\[ \frac{dy}{dx} = \frac{\cos^2 x + (1 + \sin x)\sin x}{\cos^2 x} \cdot \frac{\cos^2 x}{\cos^2 x + (1 + \sin x)^2} \]

### Step 7: Final Result
After simplification, we find that:
\[ \frac{dy}{dx} = \frac{1}{2} \]