Given the function f(x) = - (2x ^ 2 - 5) ^ 6 find f' * (x) in any form.

To find \( f'(x) \) for the function \( f(x) = - (2x^2 - 5)^6 \), we’ll need to apply the chain rule.

Given:
\[
f(x) = - (2x^2 - 5)^6
\]

Let's differentiate this step-by-step.

### Step 1: Differentiate the Outer Function
Let \( u = (2x^2 - 5) \). Then we have:
\[
f(x) = - u^6
\]

The derivative of \( -u^6 \) with respect to \( u \) is:
\[
\frac{d}{du}(-u^6) = -6u^5
\]

### Step 2: Differentiate the Inner Function
Now, we need the derivative of \( u = 2x^2 - 5 \) with respect to \( x \):
\[
\frac{du}{dx} = 4x
\]

### Step 3: Apply the Chain Rule
Now, we combine the derivatives from Steps 1 and 2 using the chain rule:
\[
f'(x) = \frac{d}{dx}[-(2x^2 - 5)^6] = -6(2x^2 - 5)^5 \cdot 4x
\]

### Step 4: Simplify
\[
f'(x) = -24x(2x^2 - 5)^5
\]

### Final Answer
\[
f'(x) = -24x(2x^2 - 5)^5
\]

Would you like further explanation on any steps?

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### Related Questions
1. How would we find the second derivative, \( f''(x) \), for this function?
2. What are the critical points of \( f(x) = -(2x^2 - 5)^6 \)?
3. How would you analyze the concavity of \( f(x) \) at specific points?
4. What is the behavior of \( f(x) \) as \( x \to \infty \) or \( x \to -\infty \)?
5. How could we find the local maxima and minima of \( f(x) \) on a closed interval?

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### Tip
When differentiating compositions of functions, use the chain rule systematically: start from the outermost function and work inward. This approach minimizes mistakes.

Learn how to find the derivative of the function f(x) = - (2x^2 - 5)^6 using the chain rule. The solution breaks down each step of the differentiation process and provides the final result f'(x) = -24x(2x^2 - 5)^5. Suitable for students in Grades 11-12 studying calculus.

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