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.