`tan^-1((2x)/(1-15x^2))`

Answer

To solve the problem of differentiating the function \( y = \tan^{-1}\left(\frac{2x}{1 - 15x^2}\right) \), we will use the formula for the difference of inverse tangents. The formula states that:

\[ \tan^{-1}\left(\frac{x - y}{1 - xy}\right) = \tan^{-1}(x) - \tan^{-1}(y) \]

### Step-by-Step Solution:

1. **Identify x and y**:
We can rewrite the expression in the form of the formula:
\[ \frac{2x}{1 - 15x^2} = \frac{5x - 3x}{1 - 5x \cdot 3x} \]
Here, we can let \( x = 5x \) and \( y = 3x \).

2. **Apply the formula**:
Using the formula, we have:
\[ y = \tan^{-1}(5x) - \tan^{-1}(3x) \]

3. **Differentiate both sides**:
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(5x)) - \frac{d}{dx}(\tan^{-1}(3x)) \]

4. **Use the derivative of arctan**:
The derivative of \( \tan^{-1}(u) \) is given by \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \). Thus,
- For \( \tan^{-1}(5x) \):
\[ \frac{d}{dx}(\tan^{-1}(5x)) = \frac{1}{1 + (5x)^2} \cdot 5 = \frac{5}{1 + 25x^2} \]
- For \( \tan^{-1}(3x) \):
\[ \frac{d}{dx}(\tan^{-1}(3x)) = \frac{1}{1 + (3x)^2} \cdot 3 = \frac{3}{1 + 9x^2} \]

5. **Combine the results**:
Now substituting back into the derivative:
\[ \frac{dy}{dx} = \frac{5}{1 + 25x^2} - \frac{3}{1 + 9x^2} \]

6. **Final expression**:
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{5}{1 + 25x^2} - \frac{3}{1 + 9x^2} \]