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The problem in the image states:

**Q46: Find the value of \(\frac{dy}{dx}\) at \(x = 2\) if \(y = \tan^{-1} \left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)\).**

### Solution:

Given: 

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

We need to find \(\frac{dy}{dx}\) at \(x = 2\).

#### Step 1: Simplify the argument of the inverse tangent.

Let:

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

Using the inverse tangent formula, \(\frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx}\).

#### Step 2: Find \(\frac{du}{dx}\).

Differentiate \(u\) with respect to \(x\):

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

Use the quotient rule:

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

Simplify the expression:

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

#### Step 3: Find \(\frac{dy}{dx}\).

Substitute \(u\) and \(\frac{du}{dx}\) into the derivative:

\[
\frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{1}{\sqrt{x}(1-\sqrt{x})^2}
\]

Since \(u = \frac{1+\sqrt{x}}{1-\sqrt{x}}\), simplify \(1+u^2\):

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

Thus:

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

#### Step 4: Evaluate at \(x = 2\).

Substitute \(x = 2\):

\[
\frac{dy}{dx} = \frac{\sqrt{2}}{2(1+2)} = \frac{\sqrt{2}}{6}
\]

### Final Answer:

\[
\frac{dy}{dx} = \frac{\sqrt{2}}{6}
\]

Would you like a detailed breakdown of any step, or do you have any other questions?

### Related Questions:
1. How do you apply the chain rule in differentiation?
2. What is the inverse tangent function and how do you differentiate it?
3. How do you simplify complex fractions in calculus problems?
4. What is the quotient rule in differentiation?
5. How do you evaluate trigonometric functions at specific points?

### Tip:
Always simplify the argument of inverse trigonometric functions to make differentiation more manageable!

Find the value of dy/dx at x = 2 if y = tan^(-1)((1+√x)/(1-√x)).

Learn how to calculate the derivative dy/dx at x = 2 for the function y = tan^(-1)((1+√x)/(1-√x)). This solution involves using the quotient rule, simplifying inverse trigonometric functions, and applying the chain rule. Suitable for college-level calculus students.

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