`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))`

Explanation

To solve the problem \( \frac{d}{dx}\left[\tan^{-1}\left(\frac{a-x}{1+ax}\right)\right] \), we can use the identity for the difference of inverse tangents. Here’s the step-by-step solution:

### Step 1: Recognize the Identity
We can use the identity:
\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A - B}{1 + AB}\right) \]
In our case, let \( A = a \) and \( B = x \). Thus, we can express our function as:
\[ \tan^{-1}\left(\frac{a-x}{1+ax}\right) = \tan^{-1}(a) - \tan^{-1}(x) \]

### Step 2: Differentiate the Expression
Now, we differentiate the expression:
\[ \frac{d}{dx}\left[\tan^{-1}(a) - \tan^{-1}(x)\right] \]
Since \( \tan^{-1}(a) \) is a constant, its derivative is 0. Therefore, we only need to differentiate \( -\tan^{-1}(x) \):
\[ \frac{d}{dx}[-\tan^{-1}(x)] = -\frac{1}{1+x^2} \]

### Step 3: Combine the Results
Thus, the derivative of the original expression is:
\[ \frac{d}{dx}\left[\tan^{-1}\left(\frac{a-x}{1+ax}\right)\right] = -\frac{1}{1+x^2} \]

### Final Answer
The final result is:
\[ -\frac{1}{1+x^2} \]

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