`tan^-1((5x)/(1-6x^2))`

Answer

To solve the problem of differentiating the function \( y = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \), we will follow these steps:

### Step 1: Rewrite the Function
Let
\[ y = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]

### Step 2: Identify the Formula
We can use the formula for the tangent of a sum:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
In our case, we can express \( \frac{5x}{1 - 6x^2} \) in terms of \( 3x \) and \( 2x \):
\[ \frac{5x}{1 - 6x^2} = \frac{3x + 2x}{1 - (3x)(2x)} \]
This means we can write:
\[ y = \tan^{-1}(3x) + \tan^{-1}(2x) \]

### Step 3: Differentiate the Function
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(3x)) + \frac{d}{dx}(\tan^{-1}(2x)) \]

Using the derivative formula for \( \tan^{-1}(u) \), which is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \):
1. For \( \tan^{-1}(3x) \):
\[ \frac{d}{dx}(\tan^{-1}(3x)) = \frac{1}{1 + (3x)^2} \cdot 3 = \frac{3}{1 + 9x^2} \]

2. For \( \tan^{-1}(2x) \):
\[ \frac{d}{dx}(\tan^{-1}(2x)) = \frac{1}{1 + (2x)^2} \cdot 2 = \frac{2}{1 + 4x^2} \]

### Step 4: Combine the Derivatives
Now, we combine the results:
\[ \frac{dy}{dx} = \frac{3}{1 + 9x^2} + \frac{2}{1 + 4x^2} \]

### Final Answer
Thus, the derivative of the function \( y = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \) is:
\[ \frac{dy}{dx} = \frac{3}{1 + 9x^2} + \frac{2}{1 + 4x^2} \]