`tan^-1""((3-5x)/(1+15x))`

Answer

To differentiate the expression \( \tan^{-1}\left(\frac{3 - 5x}{1 + 15x}\right) \), we will follow these steps:

### Step 1: Identify the expression
We start with the expression:
\[ y = \tan^{-1}\left(\frac{3 - 5x}{1 + 15x}\right) \]

### Step 2: Use the formula for the difference of arctangents
We can use the formula for the difference of arctangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
In our case, we can rewrite the expression as:
\[ y = \tan^{-1}(3) - \tan^{-1}(5x) \]
This is because:
- \( a = 3 \)
- \( b = 5x \)

### Step 3: Differentiate the expression
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\tan^{-1}(3)\right) - \frac{d}{dx}\left(\tan^{-1}(5x)\right) \]

Since \( \tan^{-1}(3) \) is a constant, its derivative is 0. Therefore, we only need to differentiate \( \tan^{-1}(5x) \):
\[ \frac{d}{dx}\left(\tan^{-1}(5x)\right) = \frac{1}{1 + (5x)^2} \cdot \frac{d}{dx}(5x) = \frac{5}{1 + 25x^2} \]

### Step 4: Combine the results
Putting it all together, we have:
\[ \frac{dy}{dx} = 0 - \frac{5}{1 + 25x^2} = -\frac{5}{1 + 25x^2} \]

### Final Result
Thus, the derivative of the given expression is:
\[ \frac{dy}{dx} = -\frac{5}{1 + 25x^2} \]

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