Answer is: option1
\( -2 \)Solution:
The Chain Rule will have to be applied three times in this example. To make it easier to visualize, we have underlined the “inner” functions to be left alone as we find the derivative. Notice how the \(2x\) is left alone until the very end.
\( P'(x) = 2(\sin(h(2x))) \cdot \cos(h(2x)) \cdot h'(2x) \cdot 2 \)
Because \(2x = 2 \left( \frac{\pi}{2} \right) = \pi\), you can write:
\( P'\left( \frac{\pi}{2} \right) = 2(\sin(h(\pi))) \cdot \cos(h(\pi)) \cdot h'(\pi) \cdot 2 \)
\( P'\left( \frac{\pi}{2} \right) = 2(\sin\left( \frac{\pi}{4} \right)) \cdot \cos\left( \frac{\pi}{4} \right) \cdot (-1) \cdot 2 \)
\( P'\left( \frac{\pi}{2} \right) = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} \cdot -2 \)
\( P'\left( \frac{\pi}{2} \right) = -2 \) Ans