If `y=sqrt((1-x)/(1+x)` then `(dy)/(dx)` equals

Answer

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \sqrt{\frac{1-x}{1+x}} \), we can follow these steps:

### Step 1: Rewrite the Function
First, we rewrite the function in a more manageable form:
\[ y = \left( \frac{1-x}{1+x} \right)^{\frac{1}{2}} \]

### Step 2: Take the Natural Logarithm
Next, we take the natural logarithm of both sides:
\[ \ln y = \frac{1}{2} \ln\left(\frac{1-x}{1+x}\right) \]

### Step 3: Differentiate Both Sides
Now, we differentiate both sides with respect to \( x \). Using the chain rule on the left side and the properties of logarithms on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\frac{1-x}{1+x}} \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right) \]

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

### Step 5: Substitute Back
Now, substituting this back into our derivative equation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1+x}{1-x} \cdot \frac{-2}{(1+x)^2} \]
This simplifies to:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{-1}{(1-x)(1+x)} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot \frac{-1}{(1-x)(1+x)} \]

### Step 7: Substitute \( y \) Back
Now substitute \( y = \sqrt{\frac{1-x}{1+x}} \) back into the equation:
\[ \frac{dy}{dx} = \sqrt{\frac{1-x}{1+x}} \cdot \frac{-1}{(1-x)(1+x)} \]

### Step 8: Simplify
This gives us:
\[ \frac{dy}{dx} = \frac{-\sqrt{1-x}}{(1-x)(1+x)\sqrt{1+x}} \]

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