If ` y = tan^(-1) ((4sqrt(x))/(1 - 4x))" then" (dy)/(dx)` is

Options

A. `(2)/(1 + 4x )`
B. `(-2)/((1 + 4x) sqrt(x))`
C. `(-2)/(1 + 4x)`
D. `(2)/((1 + 4x) sqrt(x))`

Explanation

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

### Step 1: Recognize the form of the function
We have \( y = \tan^{-1} \left( \frac{4\sqrt{x}}{1 - 4x} \right) \). This resembles the tangent double angle formula, where:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
We can express \( \frac{4\sqrt{x}}{1 - 4x} \) in terms of \( \tan(2\theta) \).

### Step 2: Set up the substitution
Let \( \tan(\theta) = 2\sqrt{x} \). Then, we can express \( \tan(2\theta) \) as follows:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} = \frac{2(2\sqrt{x})}{1 - (2\sqrt{x})^2} = \frac{4\sqrt{x}}{1 - 4x} \]
This means that:
\[ y = \tan^{-1}(\tan(2\theta)) = 2\theta \]

### Step 3: Find \( \theta \)
From our substitution, we have:
\[ \theta = \tan^{-1}(2\sqrt{x}) \]
Thus, we can rewrite \( y \) as:
\[ y = 2\tan^{-1}(2\sqrt{x}) \]

### Step 4: Differentiate \( y \)
Now, we can differentiate \( y \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx} \left( \tan^{-1}(2\sqrt{x}) \right) \]
Using the derivative of \( \tan^{-1}(u) \), which is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \), we have:
\[ \frac{d}{dx} \left( \tan^{-1}(2\sqrt{x}) \right) = \frac{1}{1 + (2\sqrt{x})^2} \cdot \frac{d}{dx}(2\sqrt{x}) \]

### Step 5: Differentiate \( 2\sqrt{x} \)
Now, we differentiate \( 2\sqrt{x} \):
\[ \frac{d}{dx}(2\sqrt{x}) = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}} \]

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

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2}{(1 + 4x)\sqrt{x}} \]