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

Answer

To differentiate the expression \( y = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ y = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]
We can express \( 5x \) as \( 3x + 2x \) so that we can use the formula for the sum of arctangents.

### Step 2: Use the Arctangent Addition Formula
Using the formula for the tangent of a sum:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
we can rewrite:
\[ y = \tan^{-1}(3x) + \tan^{-1}(2x) \]
This gives us:
\[ y = \tan^{-1}(3x) + \tan^{-1}(2x) \]

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

Using the derivative of \( \tan^{-1}(u) \) which is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \):
1. For \( \tan^{-1}(3x) \):
\[ \frac{d}{dx}\left(\tan^{-1}(3x)\right) = \frac{1}{1 + (3x)^2} \cdot 3 = \frac{3}{1 + 9x^2} \]
2. For \( \tan^{-1}(2x) \):
\[ \frac{d}{dx}\left(\tan^{-1}(2x)\right) = \frac{1}{1 + (2x)^2} \cdot 2 = \frac{2}{1 + 4x^2} \]

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

### Step 5: Final Expression
Thus, the final expression for the derivative is:
\[ \frac{dy}{dx} = \frac{3}{1 + 9x^2} + \frac{2}{1 + 4x^2} \]