To solve the given problem, let’s carefully work through the information provided in the image:

The problem is:

If $f(x) = 5 \cos^2(\pi - x)$, then $f'\left(\frac{\pi}{2}\right)$ is?

Step-by-step Solution:

1. Rewrite $f(x)$: Using the identity for cosine: $\cos(\pi - x) = -\cos(x)$ Thus: $f(x) = 5 \cos^2(\pi - x) = 5 \cos^2(-x)$ And since $\cos(-x) = \cos(x)$: $f(x) = 5 \cos^2(x)$

2. Differentiate $f(x)$: Using the chain rule: $\frac{d}{dx}[\cos^2(x)] = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x)$ So: $f'(x) = 5 \cdot (-2\cos(x)\sin(x)) = -10\cos(x)\sin(x)$

3. Simplify $f'(x)$: Using the double-angle formula: $\sin(2x) = 2\sin(x)\cos(x)$ Then: $f'(x) = -10 \cdot \frac{\sin(2x)}{2} = -5\sin(2x)$

4. Evaluate $f'\left(\frac{\pi}{2}\right)$: Substitute $x = \frac{\pi}{2}$ into $f'(x) = -5\sin(2x)$: $f'\left(\frac{\pi}{2}\right) = -5\sin\left(2 \cdot \frac{\pi}{2}\right) = -5\sin(\pi)$ Since $\sin(\pi) = 0$: $f'\left(\frac{\pi}{2}\right) = -5 \cdot 0 = 0$

Final Answer:

$\boxed{0}$

Let me know if you want more details or have any questions! Here are some related questions to explore:

1. What are the key trigonometric identities used in differentiating trigonometric functions?
2. How does the chain rule apply to composite functions like $\cos^2(x)$?
3. What is the significance of the double-angle formulas in simplifying trigonometric derivatives?
4. How does $f(x) = \cos^2(x)$ change when evaluated over different intervals?
5. Why is $\sin(\pi)$ equal to zero?

Tip: Always simplify trigonometric expressions using standard identities before differentiating to make calculations easier.