Find the derivative of the following function with respect to 'x'
`logsqrt((1+x^2)/(1-x^2))`

Answer

To find the derivative of the function \( y = \log \sqrt{\frac{1+x^2}{1-x^2}} \) with respect to \( x \), we can follow these steps:

### Step 1: Simplify the function
We start with the function:
\[ y = \log \sqrt{\frac{1+x^2}{1-x^2}} \]
Using the property of logarithms, we can rewrite this as:
\[ y = \frac{1}{2} \log \left( \frac{1+x^2}{1-x^2} \right) \]

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

### Step 3: Apply the quotient rule
To differentiate \( \frac{1+x^2}{1-x^2} \), we use the quotient rule:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( u = 1+x^2 \) and \( v = 1-x^2 \).

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\[ \frac{du}{dx} = 2x, \quad \frac{dv}{dx} = -2x \]

Now applying the quotient rule:
\[ \frac{d}{dx} \left( \frac{1+x^2}{1-x^2} \right) = \frac{(1-x^2)(2x) - (1+x^2)(-2x)}{(1-x^2)^2} \]

### Step 4: Simplify the derivative
Expanding the numerator:
\[ = \frac{2x(1-x^2) + 2x(1+x^2)}{(1-x^2)^2} \]
\[ = \frac{2x(1 - x^2 + 1 + x^2)}{(1-x^2)^2} = \frac{2x(2)}{(1-x^2)^2} = \frac{4x}{(1-x^2)^2} \]

### Step 5: Substitute back into the derivative
Now substituting back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\frac{1+x^2}{1-x^2}} \cdot \frac{4x}{(1-x^2)^2} \]
\[ = \frac{1}{2} \cdot \frac{1-x^2}{1+x^2} \cdot \frac{4x}{(1-x^2)^2} \]
\[ = \frac{2x(1-x^2)}{(1+x^2)(1-x^2)^2} \]

### Final Result
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = \frac{2x}{(1+x^2)(1-x^2)} \]