If f(x) = sin x + cos x, then fleft(frac{pi}{ | vedclass

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(B) Given $f(x) = \sin x + \cos x$. First derivative: $f'(x) = \cos x - \sin x$. Second derivative: $f''(x) = -\sin x - \cos x$. Third derivative: $f'

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$2$

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(B) Given $f(x) = \sin x + \cos x$. First derivative: $f'(x) = \cos x - \sin x$. Second derivative: $f''(x) = -\sin x - \cos x$. Third derivative: $f'''(x) = -\cos x + \sin x$. Fourth derivative: $f^{(iv)}(x) = \sin x + \cos x$. Now, calculate the value at $x = \frac{\pi}{4}$: $f\left(\frac{\pi}{4}ight) = \sin\left(\frac{\pi}{4}ight) + \cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Similarly, $f^{(iv)}\left(\frac{\pi}{4}ight) = \sin\left(\frac{\pi}{4}ight) + \cos\left(\frac{\pi}{4}ight) = \sqrt{2}$. Therefore, $f\left(\frac{\pi}{4}ight) f^{(iv)}\left(\frac{\pi}{4}ight) = \sqrt{2} 	imes \sqrt{2} = 2$.

If $f(x) = \sin x + \cos x$, then $f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right)$ is equal to

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