`sin (tan ^(-1)2x)`

Answer

To find the derivative of the function \( y = \sin(\tan^{-1}(2x)) \) with respect to \( x \), we will follow these steps:

### Step 1: Identify the function and its composition
Let:
\[ y = \sin(\tan^{-1}(2x)) \]
This function is a composition of the sine function and the inverse tangent function.

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The chain rule states that if \( y = f(g(x)) \), then:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]
In our case, let:
- \( f(u) = \sin(u) \) where \( u = \tan^{-1}(2x) \)
- \( g(x) = \tan^{-1}(2x) \)

### Step 3: Differentiate the outer function
The derivative of \( f(u) = \sin(u) \) is:
\[ f'(u) = \cos(u) \]
Thus,
\[ f'(\tan^{-1}(2x)) = \cos(\tan^{-1}(2x)) \]

### Step 4: Differentiate the inner function
Next, we differentiate \( g(x) = \tan^{-1}(2x) \):
\[ g'(x) = \frac{1}{1 + (2x)^2} \cdot \frac{d}{dx}(2x) = \frac{1}{1 + 4x^2} \cdot 2 = \frac{2}{1 + 4x^2} \]

### Step 5: Combine the derivatives
Now, we can combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \cos(\tan^{-1}(2x)) \cdot \frac{2}{1 + 4x^2} \]

### Step 6: Simplify \( \cos(\tan^{-1}(2x)) \)
To simplify \( \cos(\tan^{-1}(2x)) \), we can use the identity:
\[ \cos(\tan^{-1}(a)) = \frac{1}{\sqrt{1 + a^2}} \]
Here, \( a = 2x \), so:
\[ \cos(\tan^{-1}(2x)) = \frac{1}{\sqrt{1 + (2x)^2}} = \frac{1}{\sqrt{1 + 4x^2}} \]

### Step 7: Final expression for the derivative
Substituting this back into our derivative expression gives:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 + 4x^2}} \cdot \frac{2}{1 + 4x^2} = \frac{2}{(1 + 4x^2) \sqrt{1 + 4x^2}} \]

Thus, the final answer is:
\[ \frac{dy}{dx} = \frac{2}{(1 + 4x^2) \sqrt{1 + 4x^2}} \]