Differentiate the following w.r.t. x :
`f(x)=log[(2-x)^(1//2)(x^(2)-1)^(-1//4)]`

Answer

To differentiate the function \( f(x) = \log\left[(2-x)^{1/2}(x^2-1)^{-1/4}\right] \) with respect to \( x \), we can follow these steps:

### Step 1: Simplify the logarithmic expression
Using the properties of logarithms, we can separate the logarithm of a product into the sum of logarithms:
\[ f(x) = \log\left[(2-x)^{1/2}\right] + \log\left[(x^2-1)^{-1/4}\right] \]
This can be rewritten as:
\[ f(x) = \frac{1}{2} \log(2-x) - \frac{1}{4} \log(x^2-1) \]

### Step 2: Differentiate each term
Now, we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}\left(\frac{1}{2} \log(2-x)\right) - \frac{d}{dx}\left(\frac{1}{4} \log(x^2-1)\right) \]

Using the chain rule for differentiation:
1. For the first term:
\[ \frac{d}{dx}\left(\frac{1}{2} \log(2-x)\right) = \frac{1}{2} \cdot \frac{1}{2-x} \cdot (-1) = -\frac{1}{2(2-x)} \]

2. For the second term:
\[ \frac{d}{dx}\left(-\frac{1}{4} \log(x^2-1)\right) = -\frac{1}{4} \cdot \frac{1}{x^2-1} \cdot (2x) = -\frac{1}{2} \cdot \frac{x}{x^2-1} \]

### Step 3: Combine the derivatives
Now, we can combine the derivatives:
\[ f'(x) = -\frac{1}{2(2-x)} - \frac{x}{2(x^2-1)} \]

### Step 4: Simplify the expression
To combine these fractions, we can find a common denominator, which is \( 2(2-x)(x^2-1) \):
\[ f'(x) = -\frac{(x^2-1) + x(2-x)}{2(2-x)(x^2-1)} \]

Expanding the numerator:
\[ -(x^2 - 1 + 2x - x^2) = -(-2x + 1) = 2x - 1 \]

Thus, the derivative simplifies to:
\[ f'(x) = \frac{2x - 1}{2(2-x)(x^2-1)} \]

### Final Answer
The derivative of the function \( f(x) \) with respect to \( x \) is:
\[ f'(x) = \frac{2x - 1}{2(2-x)(x^2-1)} \]