`(d)/(dx)[cot^(-1)sqrt((x)/(1-x))]=`

Options

A. `(1)/(sqrt(x(1-x)))`
B. `(-1)/(sqrt(4x(1-x)))`
C. `(-1)/(2sqrt(x(1-x)))`
D. `(1)/(2sqrt(x(1-x^(2))))`

Answer

To solve the problem \(\frac{d}{dx} \left[ \cot^{-1} \left( \sqrt{\frac{x}{1-x}} \right) \right]\), we will use the chain rule and the derivative of the inverse cotangent function.

### Step-by-Step Solution:

1. **Identify the outer and inner functions**:
- Outer function: \( f(u) = \cot^{-1}(u) \)
- Inner function: \( g(x) = \sqrt{\frac{x}{1-x}} \)

2. **Differentiate the outer function**:
The derivative of \( f(u) = \cot^{-1}(u) \) is:
\[ f'(u) = -\frac{1}{1 + u^2} \]

3. **Differentiate the inner function**:
We need to differentiate \( g(x) = \sqrt{\frac{x}{1-x}} \). First, rewrite it:
\[ g(x) = \left( \frac{x}{1-x} \right)^{1/2} \]
Now, apply the chain rule:
\[ g'(x) = \frac{1}{2} \left( \frac{x}{1-x} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{x}{1-x} \right) \]

4. **Differentiate the fraction**:
To differentiate \( \frac{x}{1-x} \), use the quotient rule:
\[ \frac{d}{dx} \left( \frac{x}{1-x} \right) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1-x + x}{(1-x)^2} = \frac{1}{(1-x)^2} \]

5. **Combine the derivatives**:
Now substitute back into the derivative of \( g(x) \):
\[ g'(x) = \frac{1}{2} \left( \frac{x}{1-x} \right)^{-1/2} \cdot \frac{1}{(1-x)^2} \]
Simplifying gives:
\[ g'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{x}{1-x}}} \cdot \frac{1}{(1-x)^2} = \frac{1}{2} \cdot \frac{\sqrt{1-x}}{\sqrt{x}} \cdot \frac{1}{(1-x)^2} \]

6. **Apply the chain rule**:
Now, apply the chain rule:
\[ \frac{d}{dx} \left[ \cot^{-1} \left( \sqrt{\frac{x}{1-x}} \right) \right] = f'(g(x)) \cdot g'(x) \]
Substituting \( g(x) \) and \( g'(x) \):
\[ = -\frac{1}{1 + \left( \sqrt{\frac{x}{1-x}} \right)^2} \cdot g'(x) \]

7. **Simplify the expression**:
Note that:
\[ \left( \sqrt{\frac{x}{1-x}} \right)^2 = \frac{x}{1-x} \]
Thus:
\[ 1 + \frac{x}{1-x} = \frac{1-x + x}{1-x} = \frac{1}{1-x} \]
Therefore:
\[ -\frac{1}{\frac{1}{1-x}} = -(1-x) \]

8. **Final expression**:
Combine everything:
\[ \frac{d}{dx} \left[ \cot^{-1} \left( \sqrt{\frac{x}{1-x}} \right) \right] = -(1-x) \cdot \frac{1}{2} \cdot \frac{\sqrt{1-x}}{\sqrt{x}} \cdot \frac{1}{(1-x)^2} \]
Simplifying gives:
\[ = -\frac{1}{2} \cdot \frac{\sqrt{1-x}}{\sqrt{x}(1-x)} \]

### Final Answer:
\[ \frac{d}{dx} \left[ \cot^{-1} \left( \sqrt{\frac{x}{1-x}} \right) \right] = -\frac{1}{2} \cdot \frac{\sqrt{1-x}}{\sqrt{x}(1-x)} \]