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

Answer

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

### Step 1: Define the function
Let \( y = \tan^{-1}\left(\frac{4\sqrt{x}}{1 - 4x}\right) \).

### Step 2: Differentiate using the chain rule
We will use the chain rule for differentiation. The derivative of \( \tan^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{4\sqrt{x}}{1 - 4x} \).

### Step 3: Differentiate \( u \)
Now, we need to differentiate \( u \):
\[ u = \frac{4\sqrt{x}}{1 - 4x} \]
We will use the quotient rule, which states that if \( u = \frac{f(x)}{g(x)} \), then:
\[ \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
Here, \( f(x) = 4\sqrt{x} \) and \( g(x) = 1 - 4x \).

#### Step 3.1: Differentiate \( f(x) \) and \( g(x) \)
1. \( f(x) = 4\sqrt{x} \)
- \( f'(x) = 4 \cdot \frac{1}{2\sqrt{x}} = \frac{2}{\sqrt{x}} \)

2. \( g(x) = 1 - 4x \)
- \( g'(x) = -4 \)

#### Step 3.2: Apply the quotient rule
Now substituting \( f(x) \), \( f'(x) \), \( g(x) \), and \( g'(x) \) into the quotient rule:
\[ \frac{du}{dx} = \frac{\left(\frac{2}{\sqrt{x}}\right)(1 - 4x) - (4\sqrt{x})(-4)}{(1 - 4x)^2} \]
Simplifying this:
\[ \frac{du}{dx} = \frac{\frac{2(1 - 4x)}{\sqrt{x}} + 16\sqrt{x}}{(1 - 4x)^2} \]
Combine the terms in the numerator:
\[ \frac{du}{dx} = \frac{2 - 8x + 16x}{\sqrt{x}(1 - 4x)^2} = \frac{2 + 8x}{\sqrt{x}(1 - 4x)^2} \]

### Step 4: Substitute back into the derivative of \( y \)
Now substitute \( u \) and \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{4\sqrt{x}}{1 - 4x}\right)^2} \cdot \frac{2 + 8x}{\sqrt{x}(1 - 4x)^2} \]

### Step 5: Simplify the expression
First, simplify \( 1 + u^2 \):
\[ u^2 = \left(\frac{4\sqrt{x}}{1 - 4x}\right)^2 = \frac{16x}{(1 - 4x)^2} \]
Thus,
\[ 1 + u^2 = 1 + \frac{16x}{(1 - 4x)^2} = \frac{(1 - 4x)^2 + 16x}{(1 - 4x)^2} \]
Now, we can write:
\[ \frac{dy}{dx} = \frac{(1 - 4x)^2}{(1 - 4x)^2 + 16x} \cdot \frac{2 + 8x}{\sqrt{x}(1 - 4x)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2 + 8x}{\sqrt{x}((1 - 4x)^2 + 16x)} \]

### Final Answer
Thus, the derivative of \( y = \tan^{-1}\left(\frac{4\sqrt{x}}{1 - 4x}\right) \) is:
\[ \frac{dy}{dx} = \frac{2 + 8x}{\sqrt{x}((1 - 4x)^2 + 16x)} \]