QuestionIf $f(x)=\sin \left(4 x+\frac{\pi}{2}\right)$, find $f^{\prime}\left(\frac{\pi}{4}\right)$. 0 4 $-4$ 1

Studdy Solution
Evaluate $f'(x)$ at $x = \frac{\pi}{4}$.
Substitute $x = \frac{\pi}{4}$ into the derivative:
$$f'\left(\frac{\pi}{4}\right) = 4 \cos \left(4\left(\frac{\pi}{4}\right) + \frac{\pi}{2}\right)$$
Simplify the expression inside the cosine function:
$$4\left(\frac{\pi}{4}\right) = \pi$$
Thus:
$$f'\left(\frac{\pi}{4}\right) = 4 \cos \left(\pi + \frac{\pi}{2}\right)$$
$$f'\left(\frac{\pi}{4}\right) = 4 \cos \left(\frac{3\pi}{2}\right)$$
The value of $\cos \left(\frac{3\pi}{2}\right)$ is $0$.
Therefore:
$$f'\left(\frac{\pi}{4}\right) = 4 \times 0 = 0$$
The value of $f'\left(\frac{\pi}{4}\right)$ is:
$$\boxed{0}$$