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If $f(x)=\sin x+\cos x$, then $f\left(\frac{\pi}{4}\right) f^{(i v)}\left(\frac{\pi}{4}\right)$ is equal to
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2
$f(x)=\sin x+\cos x$
$f^{\prime}(x)=\cos x-\sin x$
$f^{\prime \prime}(x)=-\sin x-\cos x$
$f^{\prime \prime \prime}(x)=-\cos x+\sin x$
$f^{\prime \prime \prime \prime}(x)=\sin x+\cos x$
So, $f\left(\frac{\pi}{4}\right)=f^{\prime \prime \prime \prime}\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right)+\cos \left(\frac{\pi}{4}\right)$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
Then, $\quad f\left(\frac{\pi}{4}\right) f^{\prime \prime \prime \prime}\left(\frac{\pi}{4}\right)=\sqrt{2} \times \sqrt{2}$
$=2$
$f^{\prime}(x)=\cos x-\sin x$
$f^{\prime \prime}(x)=-\sin x-\cos x$
$f^{\prime \prime \prime}(x)=-\cos x+\sin x$
$f^{\prime \prime \prime \prime}(x)=\sin x+\cos x$
So, $f\left(\frac{\pi}{4}\right)=f^{\prime \prime \prime \prime}\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right)+\cos \left(\frac{\pi}{4}\right)$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
Then, $\quad f\left(\frac{\pi}{4}\right) f^{\prime \prime \prime \prime}\left(\frac{\pi}{4}\right)=\sqrt{2} \times \sqrt{2}$
$=2$
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