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

Options

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

Answer

To solve the problem, we need to find the derivative of the function \( y = \cot^{-1}\left(\frac{a+x}{1-ax}\right) \).

### Step-by-Step Solution:

1. **Rewrite the Function**:
We can use the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \). Thus, we can rewrite \( y \) as:
\[ y = \tan^{-1}\left(\frac{1}{\frac{a+x}{1-ax}}\right) = \tan^{-1}\left(\frac{1-ax}{a+x}\right) \]

**Hint**: Use the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \) to rewrite the function.

2. **Differentiate Using the Chain Rule**:
To differentiate \( y \) with respect to \( x \), we apply the derivative of \( \tan^{-1}(u) \), which is \( \frac{1}{1+u^2} \cdot \frac{du}{dx} \). Here, \( u = \frac{1-ax}{a+x} \).

First, we need to find \( \frac{du}{dx} \):
\[ u = \frac{1-ax}{a+x} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(a+x)(-a) - (1-ax)(1)}{(a+x)^2} \]

**Hint**: Use the quotient rule \( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \).

3. **Simplify \( \frac{du}{dx} \)**:
Simplifying the numerator:
\[ = \frac{-a(a+x) - (1-ax)}{(a+x)^2} = \frac{-a^2 - ax - 1 + ax}{(a+x)^2} = \frac{-a^2 - 1}{(a+x)^2} \]

**Hint**: Combine like terms carefully when simplifying.

4. **Substitute Back into the Derivative**:
Now substituting \( u \) and \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{1-ax}{a+x}\right)^2} \cdot \frac{-a^2 - 1}{(a+x)^2} \]

5. **Simplify the Expression**:
To simplify \( 1 + \left(\frac{1-ax}{a+x}\right)^2 \):
\[ = 1 + \frac{(1-ax)^2}{(a+x)^2} = \frac{(a+x)^2 + (1-ax)^2}{(a+x)^2} \]
Now we need to compute \( (a+x)^2 + (1-ax)^2 \):
\[ = a^2 + 2ax + x^2 + 1 - 2ax + a^2x^2 = a^2 + x^2 + 1 + a^2x^2 \]

**Hint**: Combine the terms in the numerator carefully.

6. **Final Derivative**:
Therefore, the derivative becomes:
\[ \frac{dy}{dx} = \frac{-a^2 - 1}{(a+x)^2} \cdot \frac{(a+x)^2}{a^2 + x^2 + 1 + a^2x^2} \]
Simplifying gives:
\[ \frac{dy}{dx} = \frac{-a^2 - 1}{a^2 + x^2 + 1 + a^2x^2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{-a^2 - 1}{a^2 + x^2 + 1 + a^2x^2} \]