Differentiate the following function with respect to `x` : `tan^(-1)((a+x)/(1-a x))`

Explanation

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

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

### Step 2: Use the formula for the derivative of \( \tan^{-1}(u) \)
The derivative of \( \tan^{-1}(u) \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{a+x}{1-ax} \).

### Step 3: Differentiate \( u \)
Now, we need to find \( \frac{du}{dx} \). We can use the quotient rule for differentiation:
\[ u = \frac{a+x}{1-ax} \]
Let \( f(x) = a+x \) and \( g(x) = 1-ax \). Then,
\[ \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
Calculating \( f'(x) \) and \( g'(x) \):
- \( f'(x) = 1 \)
- \( g'(x) = -a \)

Now substituting these into the quotient rule:
\[ \frac{du}{dx} = \frac{(1)(1-ax) - (a+x)(-a)}{(1-ax)^2} \]
Simplifying the numerator:
\[ = \frac{1 - ax + a(a+x)}{(1-ax)^2} = \frac{1 - ax + a^2 + ax}{(1-ax)^2} = \frac{1 + a^2}{(1-ax)^2} \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now substituting back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{a+x}{1-ax}\right)^2} \cdot \frac{1 + a^2}{(1-ax)^2} \]

### Step 5: Simplify \( 1 + u^2 \)
Now, we need to simplify \( 1 + u^2 \):
\[ u^2 = \left(\frac{a+x}{1-ax}\right)^2 = \frac{(a+x)^2}{(1-ax)^2} \]
Thus,
\[ 1 + u^2 = 1 + \frac{(a+x)^2}{(1-ax)^2} = \frac{(1-ax)^2 + (a+x)^2}{(1-ax)^2} \]
Calculating the numerator:
\[ (1-ax)^2 + (a+x)^2 = (1 - 2ax + a^2x^2) + (a^2 + 2ax + x^2) = 1 + a^2 + (a^2 + 1)x^2 \]

### Step 6: Final expression for \( \frac{dy}{dx} \)
Putting it all together:
\[ \frac{dy}{dx} = \frac{(1 + a^2)}{(1 + a^2 + (a^2 + 1)x^2)} \cdot \frac{1 + a^2}{(1-ax)^2} \]

### Final Answer
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = \frac{(1 + a^2)^2}{(1 + a^2 + (a^2 + 1)x^2)(1-ax)^2} \]