If `y=tan^(-1) ((sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))), x^2 le 1` , then find `(dy)/(dx)`

Answer

To find \(\frac{dy}{dx}\) for the given function

\[ y = \tan^{-1} \left( \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right), \quad x^2 \leq 1 \]

we will follow these steps:

### Step 1: Simplify the Argument of the Arctangent
Let

\[ u = \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \]

### Step 2: Differentiate \(y\) with Respect to \(u\)
Using the derivative of the arctangent function, we have:

\[ \frac{dy}{du} = \frac{1}{1+u^2} \]

### Step 3: Differentiate \(u\) with Respect to \(x\)
Now we need to find \(\frac{du}{dx}\). We will use the quotient rule:

\[ \frac{du}{dx} = \frac{(\sqrt{1+x^2} - \sqrt{1-x^2}) \frac{d}{dx}(\sqrt{1+x^2} + \sqrt{1-x^2}) - (\sqrt{1+x^2} + \sqrt{1-x^2}) \frac{d}{dx}(\sqrt{1+x^2} - \sqrt{1-x^2})}{(\sqrt{1+x^2} - \sqrt{1-x^2})^2} \]

### Step 4: Find the Derivatives of the Numerators
Now we compute the derivatives:

1. For \(\sqrt{1+x^2}\):
\[ \frac{d}{dx}(\sqrt{1+x^2}) = \frac{x}{\sqrt{1+x^2}} \]

2. For \(\sqrt{1-x^2}\):
\[ \frac{d}{dx}(\sqrt{1-x^2}) = \frac{-x}{\sqrt{1-x^2}} \]

### Step 5: Substitute Back into \(\frac{du}{dx}\)
Substituting these derivatives back gives:

\[ \frac{du}{dx} = \frac{(\sqrt{1+x^2} - \sqrt{1-x^2}) \left( \frac{x}{\sqrt{1+x^2}} + \frac{x}{\sqrt{1-x^2}} \right) - (\sqrt{1+x^2} + \sqrt{1-x^2}) \left( \frac{x}{\sqrt{1+x^2}} - \frac{x}{\sqrt{1-x^2}} \right)}{(\sqrt{1+x^2} - \sqrt{1-x^2})^2} \]

### Step 6: Combine the Derivatives
Now, we can combine the derivatives using the chain rule:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx} \]

### Step 7: Final Expression
Substituting \(u\) back into the equation will give us the final expression for \(\frac{dy}{dx}\).