We are tasked with finding the derivative of the function $f(x) = \sqrt{x + \sqrt{x - 3}}$.

Steps:

1. Rewrite the function for easier differentiation: $f(x) = \left( x + \sqrt{x - 3} \right)^{\frac{1}{2}}$

2. Apply the chain rule: We have an outer function $u^{\frac{1}{2}}$, where $u = x + \sqrt{x - 3}$. The derivative of $u^{\frac{1}{2}}$ is: $\frac{d}{dx} \left( u^{\frac{1}{2}} \right) = \frac{1}{2} u^{-\frac{1}{2}} \cdot \frac{du}{dx}$

3. Differentiate $u = x + \sqrt{x - 3}$: The derivative of $u$ is: $\frac{du}{dx} = \frac{d}{dx} \left( x + \sqrt{x - 3} \right) = 1 + \frac{1}{2} (x - 3)^{-\frac{1}{2}} \cdot 1$ This simplifies to: $\frac{du}{dx} = 1 + \frac{1}{2\sqrt{x - 3}}$

4. Substitute $\frac{du}{dx}$ into the chain rule: Now we substitute back into the chain rule expression: $f'(x) = \frac{1}{2} \left( x + \sqrt{x - 3} \right)^{-\frac{1}{2}} \cdot \left( 1 + \frac{1}{2\sqrt{x - 3}} \right)$

Final derivative:

Thus, the derivative of $f(x) = \sqrt{x + \sqrt{x - 3}}$ is: $f'(x) = \frac{1}{2\sqrt{x + \sqrt{x - 3}}} \cdot \left( 1 + \frac{1}{2\sqrt{x - 3}} \right)$

Would you like a breakdown of any specific part of the solution?

Related Questions:

1. How do we handle derivatives involving multiple square roots?
2. What is the geometric interpretation of the derivative of square root functions?
3. How does the chain rule apply when differentiating nested functions?
4. What other techniques can simplify derivatives involving roots?
5. Can we find the second derivative of this function?

Tip:

When dealing with nested functions, always apply the chain rule carefully, breaking the problem down into simpler pieces.