To find the derivative of the function $f(x) = x^{2/3} + \sqrt{x}(x^2 + 4)$, we will use the rules of differentiation including the power rule and the product rule.

Step 1: Differentiate each term separately.

Term 1: $x^{2/3}$

The derivative of $x^{2/3}$ is found using the power rule: $\frac{d}{dx}\left( x^{2/3} \right) = \frac{2}{3}x^{-1/3}$

Term 2: $\sqrt{x}(x^2 + 4)$

This is a product of two functions: $\sqrt{x} = x^{1/2}$ and $(x^2 + 4)$.

Using the product rule: $\frac{d}{dx}\left( u(x)v(x) \right) = u'(x)v(x) + u(x)v'(x)$ where $u(x) = x^{1/2}$ and $v(x) = x^2 + 4$.

1. Differentiate $u(x) = x^{1/2}$: $u'(x) = \frac{1}{2}x^{-1/2}$

2. Differentiate $v(x) = x^2 + 4$: $v'(x) = 2x$

Now apply the product rule: $\frac{d}{dx}\left( \sqrt{x}(x^2 + 4) \right) = \left( \frac{1}{2}x^{-1/2} \right)(x^2 + 4) + \left( x^{1/2} \right)(2x)$

Simplifying this: $= \frac{x^2 + 4}{2\sqrt{x}} + 2x^{3/2}$

Step 2: Combine the results

Now we combine the derivatives of both terms:

$f'(x) = \frac{2}{3}x^{-1/3} + \frac{x^2 + 4}{2\sqrt{x}} + 2x^{3/2}$

Final simplified form:

Thus, the derivative of the function $f(x) = x^{2/3} + \sqrt{x}(x^2 + 4)$ is:

$f'(x) = \frac{2}{3}x^{-1/3} + \frac{x^2 + 4}{2\sqrt{x}} + 2x^{3/2}$

Would you like me to go into more detail about any specific step?

Here are 5 questions related to this topic:

1. How do you apply the product rule when differentiating?
2. What is the power rule for differentiation, and when is it used?
3. How do you simplify expressions with negative or fractional exponents?
4. Can you explain why the chain rule wasn't needed here?
5. How can we interpret the result geometrically as the slope of the tangent line?

Tip: When differentiating products, always remember to apply the product rule before simplifying.