`tan ^-1((2x)/(1-x^2))`

Answer

To find the differential coefficient of the function \( f(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), we can use a substitution and some trigonometric identities. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( x = \tan(\theta) \). Then, we have:
\[ \theta = \tan^{-1}(x) \]

### Step 2: Express \( f(x) \) in terms of \( \theta \)
Using the double angle formula for tangent, we know:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Substituting \( \tan(\theta) = x \), we get:
\[ \tan(2\theta) = \frac{2x}{1 - x^2} \]
Thus, we can rewrite \( f(x) \):
\[ f(x) = \tan^{-1}\left(\tan(2\theta)\right) = 2\theta \]

### Step 3: Relate \( \theta \) back to \( x \)
Since \( \theta = \tan^{-1}(x) \), we have:
\[ f(x) = 2\tan^{-1}(x) \]

### Step 4: Differentiate \( f(x) \)
Now, we differentiate \( f(x) \) with respect to \( x \):
\[ \frac{d}{dx} f(x) = \frac{d}{dx} (2\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \), which is \( \frac{1}{1+x^2} \), we get:
\[ \frac{d}{dx} f(x) = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

### Final Result
Thus, the differential coefficient of the function \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) is:
\[ \frac{d}{dx} \left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) = \frac{2}{1+x^2} \]

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