`(d)/(dx)[tan^(-1)((a-x)/(1+ax))]` is equal to

Options

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

Answer

To solve the problem \(\frac{d}{dx}\left[\tan^{-1}\left(\frac{a-x}{1+ax}\right)\right]\), we will use the chain rule and the properties of the inverse tangent function.

### Step-by-Step Solution:

1. **Identify the function**: We need to differentiate the function \(y = \tan^{-1}\left(\frac{a-x}{1+ax}\right)\).

2. **Apply the chain rule**: The derivative of \(\tan^{-1}(u)\) with respect to \(x\) is given by \(\frac{1}{1+u^2} \cdot \frac{du}{dx}\), where \(u = \frac{a-x}{1+ax}\).

3. **Find \(u\)**:
\[ u = \frac{a-x}{1+ax} \]

4. **Differentiate \(u\)**: We need to find \(\frac{du}{dx}\).
- Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1+ax)(-1) - (a-x)(a)}{(1+ax)^2} \]
- Simplifying the numerator:
\[ = \frac{-(1+ax) + a(a-x)}{(1+ax)^2} \]
\[ = \frac{-1 - ax + a^2 - ax}{(1+ax)^2} \]
\[ = \frac{a^2 - 1 - 2ax}{(1+ax)^2} \]

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

6. **Simplify \(1 + u^2\)**:
\[ 1 + \left(\frac{a-x}{1+ax}\right)^2 = 1 + \frac{(a-x)^2}{(1+ax)^2} \]
\[ = \frac{(1+ax)^2 + (a-x)^2}{(1+ax)^2} \]

7. **Combine the expressions**:
\[ \frac{dy}{dx} = \frac{(a^2 - 1 - 2ax)}{(1 + ax)^2} \cdot \frac{(1 + ax)^2}{(1 + ax)^2 + (a-x)^2} \]

8. **Final expression**:
\[ \frac{dy}{dx} = \frac{a^2 - 1 - 2ax}{(1 + ax)^2 + (a-x)^2} \]