If `y =tan ^(-1) ((sqrt( a) -sqrt(x)) /( 1+sqrt( ax)) ) ,then (dy)/(dx)=`

Options

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

Answer

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

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

we can use the identity for the difference of inverse tangents:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left( \frac{a - b}{1 + ab} \right). \]

### Step 1: Rewrite the function using the identity
We can express \( y \) as:

\[ y = \tan^{-1}(\sqrt{a}) - \tan^{-1}(\sqrt{x}). \]

### Step 2: Differentiate using the chain rule
Now, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}(\sqrt{a}) \right) - \frac{d}{dx} \left( \tan^{-1}(\sqrt{x}) \right). \]

Since \( \tan^{-1}(\sqrt{a}) \) is a constant (as \( a \) is a constant), its derivative is 0:

\[ \frac{dy}{dx} = 0 - \frac{d}{dx} \left( \tan^{-1}(\sqrt{x}) \right). \]

### Step 3: Apply the derivative formula for inverse tangent
The derivative of \( \tan^{-1}(u) \) is given by:

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}. \]

Here, \( u = \sqrt{x} \), so we first need to find \( \frac{du}{dx} \):

\[ u = \sqrt{x} \implies \frac{du}{dx} = \frac{1}{2\sqrt{x}}. \]

### Step 4: Substitute into the derivative formula
Now substituting \( u \) back into the derivative formula:

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

### Step 5: Simplify the expression
Since \( (\sqrt{x})^2 = x \), we can simplify:

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

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:

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