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

Answer

To find the derivative of the function \( y = \sin(\tan^{-1}(2x)) \), we will apply the chain rule of differentiation step by step.

### Step-by-Step Solution:

1. **Define the function:**
Let \( y = \sin(\tan^{-1}(2x)) \).

2. **Identify the inner function:**
Let \( u = \tan^{-1}(2x) \). Therefore, we can rewrite \( y \) as:
\[ y = \sin(u) \]

3. **Differentiate using the chain rule:**
According to the chain rule, the derivative \( \frac{dy}{dx} \) can be expressed as:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

4. **Differentiate \( y \) with respect to \( u \):**
The derivative of \( \sin(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = \cos(u) \]

5. **Differentiate \( u \) with respect to \( x \):**
Now we need to find \( \frac{du}{dx} \). The derivative of \( \tan^{-1}(x) \) is given by:
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \]
Thus, for \( u = \tan^{-1}(2x) \):
\[ \frac{du}{dx} = \frac{1}{1+(2x)^2} \cdot \frac{d}{dx}(2x) = \frac{1}{1+4x^2} \cdot 2 = \frac{2}{1+4x^2} \]

6. **Combine the derivatives:**
Now substituting back into the chain rule:
\[ \frac{dy}{dx} = \cos(u) \cdot \frac{du}{dx} = \cos(\tan^{-1}(2x)) \cdot \frac{2}{1+4x^2} \]

7. **Simplify \( \cos(\tan^{-1}(2x)) \):**
To find \( \cos(\tan^{-1}(2x)) \), we can use the identity:
\[ \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1+x^2}} \]
Therefore,
\[ \cos(\tan^{-1}(2x)) = \frac{1}{\sqrt{1+(2x)^2}} = \frac{1}{\sqrt{1+4x^2}} \]

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

### Final Answer:
\[ \frac{dy}{dx} = \frac{2}{(1+4x^2)\sqrt{1+4x^2}} \]