Differentiate the following w.r.t.x. `sqrt(4+sqrt(4+sqrt(4+x^(2))))`

Answer

To differentiate the function \( y = \sqrt{4 + \sqrt{4 + \sqrt{4 + x^2}}} \) with respect to \( x \), we will use the chain rule and the formula for the derivative of the square root function.

### Step-by-Step Solution:

1. **Define the function**:
\[ y = \sqrt{4 + \sqrt{4 + \sqrt{4 + x^2}}} \]

2. **Differentiate both sides with respect to \( x \)**:
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{4 + \sqrt{4 + \sqrt{4 + x^2}}}} \cdot \frac{d}{dx}(4 + \sqrt{4 + \sqrt{4 + x^2}}) \]

3. **Differentiate the inner function**:
The derivative of \( 4 \) is \( 0 \), so we focus on \( \sqrt{4 + \sqrt{4 + x^2}} \):
\[ \frac{d}{dx}(4 + \sqrt{4 + \sqrt{4 + x^2}}) = \frac{1}{2\sqrt{4 + \sqrt{4 + x^2}}} \cdot \frac{d}{dx}(4 + \sqrt{4 + x^2}) \]

4. **Differentiate the next inner function**:
Again, the derivative of \( 4 \) is \( 0 \), so we differentiate \( \sqrt{4 + x^2} \):
\[ \frac{d}{dx}(4 + \sqrt{4 + x^2}) = \frac{1}{2\sqrt{4 + x^2}} \cdot \frac{d}{dx}(4 + x^2) = \frac{1}{2\sqrt{4 + x^2}} \cdot 2x = \frac{x}{\sqrt{4 + x^2}} \]

5. **Combine the derivatives**:
Now substituting back, we have:
\[ \frac{d}{dx}(4 + \sqrt{4 + \sqrt{4 + x^2}}) = \frac{1}{2\sqrt{4 + \sqrt{4 + x^2}}} \cdot \frac{x}{\sqrt{4 + x^2}} \]

6. **Substituting back into the derivative of \( y \)**:
Now we substitute this back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{4 + \sqrt{4 + \sqrt{4 + x^2}}}} \cdot \frac{1}{2\sqrt{4 + \sqrt{4 + x^2}}} \cdot \frac{x}{\sqrt{4 + x^2}} \]

7. **Final expression**:
This simplifies to:
\[ \frac{dy}{dx} = \frac{x}{4 \sqrt{4 + \sqrt{4 + \sqrt{4 + x^2}}} \cdot \sqrt{4 + \sqrt{4 + x^2}} \cdot \sqrt{4 + x^2}} \]

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