Differentiate the following w.r.t. x :
`tan^(-1)((sqrt(1+a^(2)x^(2))-1)/(ax))`

Answer

To differentiate the function \( y = \tan^{-1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right) \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the function
Let:
\[ y = \tan^{-1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right) \]

### Step 2: Use substitution
To simplify the differentiation, we can use the substitution \( ax = \tan(\theta) \). Thus, we have:
\[ x = \frac{\tan(\theta)}{a} \]

### Step 3: Substitute in the equation
Now substituting \( ax \) in the equation:
\[ y = \tan^{-1} \left( \frac{\sqrt{1 + \tan^2(\theta)} - 1}{\tan(\theta)} \right) \]
Using the identity \( \sqrt{1 + \tan^2(\theta)} = \sec(\theta) \), we can rewrite:
\[ y = \tan^{-1} \left( \frac{\sec(\theta) - 1}{\tan(\theta)} \right) \]

### Step 4: Simplify the expression
Now we can express \( \sec(\theta) \) and \( \tan(\theta) \) in terms of sine and cosine:
\[ y = \tan^{-1} \left( \frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}} \right) = \tan^{-1} \left( \frac{1 - \cos(\theta)}{\sin(\theta)} \right) \]

### Step 5: Use half-angle identity
Using the half-angle identity, we can express \( 1 - \cos(\theta) \) as:
\[ 1 - \cos(\theta) = 2 \sin^2\left(\frac{\theta}{2}\right) \]
Thus,
\[ y = \tan^{-1} \left( \frac{2 \sin^2\left(\frac{\theta}{2}\right)}{\sin(\theta)} \right) \]

### Step 6: Further simplification
Using the identity \( \sin(\theta) = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) \):
\[ y = \tan^{-1} \left( \frac{2 \sin^2\left(\frac{\theta}{2}\right)}{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)} \right) = \tan^{-1} \left( \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} \right) = \tan^{-1}(\tan\left(\frac{\theta}{2}\right)) \]

### Step 7: Final expression for \( y \)
Thus, we have:
\[ y = \frac{\theta}{2} \]
Since \( \theta = \tan^{-1}(ax) \), we can write:
\[ y = \frac{1}{2} \tan^{-1}(ax) \]

### Step 8: Differentiate \( y \) with respect to \( x \)
Now we differentiate:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{d}{dx} \left( \tan^{-1}(ax) \right) \]
Using the derivative of \( \tan^{-1}(u) \):
\[ \frac{d}{dx} \left( \tan^{-1}(u) \right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = ax \), thus \( \frac{du}{dx} = a \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1 + (ax)^2} \cdot a = \frac{a}{2(1 + a^2 x^2)} \]

### Final Answer
The derivative of the given function with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{a}{2(1 + a^2 x^2)} \]