(v) Let `y= sqrt{(1+x)/(1-x)}` find dy/dx

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \sqrt{\frac{1+x}{1-x}}\), we will use the quotient rule and the chain rule of differentiation. Let's go through the steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \sqrt{\frac{1+x}{1-x}} \]
This can be rewritten as:
\[ y = \left( \frac{1+x}{1-x} \right)^{1/2} \]

### Step 2: Use the chain rule
To differentiate \(y\), we apply the chain rule. The derivative of \(y\) with respect to \(x\) is:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1+x}{1-x} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{1+x}{1-x} \right) \]

### Step 3: Differentiate the inner function using the quotient rule
Now we need to differentiate \(\frac{1+x}{1-x}\) using the quotient rule:
\[ \frac{d}{dx} \left( \frac{1+x}{1-x} \right) = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} \]
Simplifying this gives:
\[ \frac{(1-x) + (1+x)}{(1-x)^2} = \frac{2}{(1-x)^2} \]

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

### Step 5: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{(1-x)^2} \cdot \left( \frac{1+x}{1-x} \right)^{-1/2} \]
We can rewrite \(\left( \frac{1+x}{1-x} \right)^{-1/2}\) as:
\[ \frac{1}{\sqrt{\frac{1+x}{1-x}}} = \frac{\sqrt{1-x}}{\sqrt{1+x}} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{\sqrt{1-x}}{(1-x)^2 \sqrt{1+x}} \]

### Final Answer
So, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\sqrt{1-x}}{(1-x)^2 \sqrt{1+x}} \]