If `y=tan^(-1) ((cos x)/(1-sin x)),"then " dy/dx=`

Options

A. `(-3)/(2)`
B. `(3)/(2)`
C. `(-1)/(2)`
D. `(1)/(2)`

Answer

To find the derivative of the function \( y = \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) \), we will use the chain rule and the quotient rule. Here’s a step-by-step solution:

### Step 1: Differentiate \( y \) using the chain rule
The derivative of \( y = \tan^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{\cos x}{1 - \sin x} \).

### Step 2: Find \( u^2 \)
First, we need to compute \( u^2 \):
\[ u^2 = \left( \frac{\cos x}{1 - \sin x} \right)^2 = \frac{\cos^2 x}{(1 - \sin x)^2} \]

### Step 3: Substitute \( u^2 \) into the derivative formula
Now, substituting \( u^2 \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{1 + \frac{\cos^2 x}{(1 - \sin x)^2}} \cdot \frac{du}{dx} \]

### Step 4: Simplify the expression
To simplify \( 1 + u^2 \):
\[ 1 + u^2 = 1 + \frac{\cos^2 x}{(1 - \sin x)^2} = \frac{(1 - \sin x)^2 + \cos^2 x}{(1 - \sin x)^2} \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ (1 - \sin x)^2 + \cos^2 x = 1 - 2\sin x + \sin^2 x + \cos^2 x = 1 - 2\sin x + 1 = 2 - 2\sin x \]
Thus,
\[ 1 + u^2 = \frac{2(1 - \sin x)}{(1 - \sin x)^2} \]

### Step 5: Differentiate \( u \) using the quotient rule
Now we need to differentiate \( u \):
\[ u = \frac{\cos x}{1 - \sin x} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1 - \sin x)(-\sin x) - \cos x(-\cos x)}{(1 - \sin x)^2} \]
This simplifies to:
\[ \frac{du}{dx} = \frac{-\sin x + \cos^2 x + \sin x \cos x}{(1 - \sin x)^2} = \frac{\cos^2 x + \sin x \cos x}{(1 - \sin x)^2} \]

### Step 6: Substitute \( \frac{du}{dx} \) back into the derivative formula
Now substituting \( \frac{du}{dx} \) back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\frac{2(1 - \sin x)}{(1 - \sin x)^2}} \cdot \frac{\cos^2 x + \sin x \cos x}{(1 - \sin x)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{(1 - \sin x)^2}{2(1 - \sin x)} \cdot \frac{\cos^2 x + \sin x \cos x}{(1 - \sin x)^2} \]
Canceling \( (1 - \sin x)^2 \):
\[ \frac{dy}{dx} = \frac{\cos^2 x + \sin x \cos x}{2(1 - \sin x)} \]

### Final Step: Simplify the expression
Using the identity \( \cos^2 x + \sin^2 x = 1 \):
\[ \frac{dy}{dx} = \frac{1 - \sin^2 x + \sin x \cos x}{2(1 - \sin x)} = \frac{1 - \sin x}{2(1 - \sin x)} = \frac{1}{2} \]

Thus, the final answer is:
\[ \frac{dy}{dx} = \frac{1}{2} \]