https://file.aiquickdraw.com/mathbot/file/1727493467966fyuj1ffq.jpg

The function provided in the image is:

\[
f(x) = 8x^4\sqrt{x} + \frac{4}{x^3\sqrt{x}}.
\]

To find the derivative \( f'(x) \), we need to rewrite the function in a more suitable form for differentiation and then apply the power rule and quotient rule where necessary.

### Step 1: Rewrite the function

1. The square root of \(x\), \(\sqrt{x}\), can be written as \(x^{1/2}\).
2. Now, express each term in a form suitable for the power rule.

\[
f(x) = 8x^4 \cdot x^{1/2} + 4x^{-3} \cdot x^{-1/2}.
\]

Simplify each term:
\[
f(x) = 8x^{4 + 1/2} + 4x^{-3 - 1/2}.
\]
\[
f(x) = 8x^{9/2} + 4x^{-7/2}.
\]

### Step 2: Differentiate each term

Using the power rule \(\frac{d}{dx}(x^n) = n x^{n-1}\), differentiate each term:

1. The derivative of \( 8x^{9/2} \) is:
\[
\frac{d}{dx}\left(8x^{9/2}\right) = 8 \cdot \frac{9}{2} x^{9/2 - 1} = 36x^{7/2}.
\]

2. The derivative of \( 4x^{-7/2} \) is:
\[
\frac{d}{dx}\left(4x^{-7/2}\right) = 4 \cdot \left(-\frac{7}{2}\right) x^{-7/2 - 1} = -14x^{-9/2}.
\]

### Step 3: Combine the results

The derivative is:
\[
f'(x) = 36x^{7/2} - 14x^{-9/2}.
\]

This is the final answer.

---

Would you like any further details or clarifications? Here are some follow-up questions:

1. How is the power rule applied to fractional exponents?
2. How can we simplify expressions with negative exponents?
3. What is the general rule for differentiating functions involving square roots?
4. How would the quotient rule apply if the function were written differently?
5. Can this result be simplified further by factoring?

**Tip**: When working with roots and exponents, converting them into fractional powers often simplifies the differentiation process significantly.

Let f(x) = 8x^4√x + 4 / (x^3√x). Find f'(x).

Find the derivative of the function f(x) = 8x^4√x + 4 / (x^3√x) by applying the power rule for differentiation and rewriting square roots as fractional exponents. Learn the step-by-step solution involving exponentiation and the power rule, suitable for college-level calculus students.

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