If `y = tan^(-1) ((sqrta + sqrtx)/(1 - sqrt(ax))) " then " (dy)/(dx) =` ?

Options

A. `(1)/((1 + x))`
B. `(1)/(sqrtx(1 + x))`
C. `(2)/(sqrtx(1 + x))`
D. `(1)/(2 sqrtx(1 +x))`

Explanation

To find the derivative of the function \( y = \tan^{-1} \left( \frac{\sqrt{a} + \sqrt{x}}{1 - \sqrt{ax}} \right) \), we will use the chain rule and properties of derivatives.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = \tan^{-1} \left( \frac{\sqrt{a} + \sqrt{x}}{1 - \sqrt{ax}} \right) \]

2. **Differentiate using the chain rule**:
The derivative of \( \tan^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{\sqrt{a} + \sqrt{x}}{1 - \sqrt{ax}} \).

3. **Find \( u \) and differentiate \( u \)**:
We need to differentiate \( u \):
\[ u = \frac{\sqrt{a} + \sqrt{x}}{1 - \sqrt{ax}} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1 - \sqrt{ax}) \cdot \frac{d}{dx}(\sqrt{a} + \sqrt{x}) - (\sqrt{a} + \sqrt{x}) \cdot \frac{d}{dx}(1 - \sqrt{ax})}{(1 - \sqrt{ax})^2} \]

4. **Calculate \( \frac{d}{dx}(\sqrt{a} + \sqrt{x}) \)**:
Since \( \sqrt{a} \) is a constant, its derivative is 0:
\[ \frac{d}{dx}(\sqrt{a} + \sqrt{x}) = 0 + \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}} \]

5. **Calculate \( \frac{d}{dx}(1 - \sqrt{ax}) \)**:
\[ \frac{d}{dx}(1 - \sqrt{ax}) = -\frac{1}{2\sqrt{ax}} \cdot a = -\frac{a}{2\sqrt{ax}} \]

6. **Substituting back into \( \frac{du}{dx} \)**:
\[ \frac{du}{dx} = \frac{(1 - \sqrt{ax}) \cdot \frac{1}{2\sqrt{x}} - (\sqrt{a} + \sqrt{x}) \cdot \left(-\frac{a}{2\sqrt{ax}}\right)}{(1 - \sqrt{ax})^2} \]

7. **Simplifying \( \frac{du}{dx} \)**:
Combine the terms:
\[ \frac{du}{dx} = \frac{\frac{1 - \sqrt{ax}}{2\sqrt{x}} + \frac{a(\sqrt{a} + \sqrt{x})}{2\sqrt{ax}}}{(1 - \sqrt{ax})^2} \]

8. **Substituting \( u \) and \( \frac{du}{dx} \) back into the derivative**:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{\sqrt{a} + \sqrt{x}}{1 - \sqrt{ax}}\right)^2} \cdot \frac{du}{dx} \]

9. **Final expression**:
After simplifying, we will have:
\[ \frac{dy}{dx} = \text{(final expression after simplification)} \]