find derivative: f
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We are tasked with finding the derivative of the function:

\[
f(x) = \left( x^6 - 9\sqrt{x} \right)^{14x}
\]

This involves applying the chain rule and power rule, as well as handling the product of \( 14x \) as an exponent. To differentiate such a function, it’s often easier to take the natural logarithm of both sides first, simplify, and then differentiate.

### Step-by-Step:

1. **Rewrite the Function for Clarity:**
   
   The function \( f(x) \) is in the form of \( (g(x))^{h(x)} \), where:
   \[
   g(x) = x^6 - 9\sqrt{x}, \quad h(x) = 14x
   \]

2. **Take the Natural Logarithm of Both Sides:**

   \[
   \ln(f(x)) = \ln\left( (x^6 - 9\sqrt{x})^{14x} \right)
   \]

   Using the logarithmic identity \( \ln(a^b) = b\ln(a) \), we get:
   \[
   \ln(f(x)) = 14x \cdot \ln(x^6 - 9\sqrt{x})
   \]

3. **Differentiate Implicitly:**

   Now differentiate both sides with respect to \( x \), using the product rule and the chain rule.

   \[
   \frac{d}{dx}\left( \ln(f(x)) \right) = \frac{d}{dx} \left( 14x \cdot \ln(x^6 - 9\sqrt{x}) \right)
   \]

   The left-hand side:
   \[
   \frac{d}{dx} \left( \ln(f(x)) \right) = \frac{1}{f(x)} \cdot f'(x)
   \]

   The right-hand side requires the product rule:
   \[
   \frac{d}{dx} \left( 14x \cdot \ln(x^6 - 9\sqrt{x}) \right) = 14 \ln(x^6 - 9\sqrt{x}) + 14x \cdot \frac{1}{x^6 - 9\sqrt{x}} \cdot \frac{d}{dx} \left( x^6 - 9\sqrt{x} \right)
   \]

4. **Differentiate Inside \( \ln(x^6 - 9\sqrt{x}) \):**

   To compute \( \frac{d}{dx} \left( x^6 - 9\sqrt{x} \right) \), apply the power rule and chain rule:
   \[
   \frac{d}{dx} \left( x^6 - 9\sqrt{x} \right) = 6x^5 - \frac{9}{2\sqrt{x}}
   \]

5. **Combine the Results:**

   Now, substitute everything back:
   \[
   \frac{1}{f(x)} \cdot f'(x) = 14 \ln(x^6 - 9\sqrt{x}) + 14x \cdot \frac{1}{x^6 - 9\sqrt{x}} \cdot \left( 6x^5 - \frac{9}{2\sqrt{x}} \right)
   \]

6. **Multiply by \( f(x) \):**

   Finally, multiply both sides of the equation by \( f(x) \) to isolate \( f'(x) \):
   \[
   f'(x) = f(x) \left( 14 \ln(x^6 - 9\sqrt{x}) + 14x \cdot \frac{1}{x^6 - 9\sqrt{x}} \cdot \left( 6x^5 - \frac{9}{2\sqrt{x}} \right) \right)
   \]

   And substitute \( f(x) \) back:
   \[
   f'(x) = \left( x^6 - 9\sqrt{x} \right)^{14x} \left( 14 \ln(x^6 - 9\sqrt{x}) + 14x \cdot \frac{1}{x^6 - 9\sqrt{x}} \cdot \left( 6x^5 - \frac{9}{2\sqrt{x}} \right) \right)
   \]

This is the derivative \( f'(x) \).

---

Would you like me to go over the steps in more detail or clarify any part of this? 

Here are some related questions to explore further:

1. How do the chain rule and product rule interact when differentiating such functions?
2. Can we simplify this expression further for specific values of \( x \)?
3. What are some other examples of functions involving products of powers and logarithms?
4. How would you apply this method to more complicated functions with multiple exponents?
5. Can we verify this derivative with an alternative approach like implicit differentiation or logarithmic differentiation?

**Tip:** When you encounter an expression with a power and a function inside it, taking the natural logarithm first can simplify the differentiation process significantly.

find derivative: f(x) = (x^6 - 9√x)^(14x)

Learn how to differentiate the function f(x) = (x^6 - 9√x)^(14x) using chain and product rules. This detailed step-by-step solution covers logarithmic differentiation and exponentiation. Suitable for high school students in Grades 11-12.

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