Differentiate `tan^-1((4sqrt(x))/(1-4x))`

Answer

To differentiate the function \( y = \tan^{-1}\left(\frac{4\sqrt{x}}{1-4x}\right) \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression inside the inverse tangent function. We can express \( \frac{4\sqrt{x}}{1-4x} \) in terms of a tangent function.

Let:
\[ \tan(\theta) = 2\sqrt{x} \]
Then:
\[ \frac{4\sqrt{x}}{1-4x} = \tan(2\theta) \]

### Step 2: Use the identity for tangent of double angle
Using the double angle formula for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Substituting \( \tan(\theta) = 2\sqrt{x} \):
\[ \tan(2\theta) = \frac{2(2\sqrt{x})}{1 - (2\sqrt{x})^2} = \frac{4\sqrt{x}}{1 - 4x} \]

### Step 3: Rewrite the function
Thus, we can rewrite our function as:
\[ y = \tan^{-1}(\tan(2\theta)) = 2\theta \]
This implies:
\[ y = 2\tan^{-1}(2\sqrt{x}) \]

### Step 4: Differentiate the function
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx} \left( \tan^{-1}(2\sqrt{x}) \right) \]

Using the derivative of \( \tan^{-1}(u) \), where \( u = 2\sqrt{x} \):
\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

### Step 5: Calculate \( \frac{du}{dx} \)
First, we find \( \frac{du}{dx} \):
\[ u = 2\sqrt{x} \implies \frac{du}{dx} = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}} \]

### Step 6: Substitute back into the derivative
Now substituting back:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + (2\sqrt{x})^2} \cdot \frac{1}{\sqrt{x}} = 2 \cdot \frac{1}{1 + 4x} \cdot \frac{1}{\sqrt{x}} \]

### Step 7: Simplify the expression
Thus, we have:
\[ \frac{dy}{dx} = \frac{2}{\sqrt{x}(1 + 4x)} \]

### Final Answer
The derivative of \( y = \tan^{-1}\left(\frac{4\sqrt{x}}{1-4x}\right) \) is:
\[ \frac{dy}{dx} = \frac{2}{\sqrt{x}(1 + 4x)} \]