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Q.
The maximum value of the function $\sin x+\cos x$ is
Application of Derivatives
Solution:
Let $f(x)=\sin x+\cos x \Rightarrow f^{\prime}(x)=\cos x-\sin x$
and $f^{\prime \prime}(x)=-\sin x-\cos x=-(\sin x+\cos x)$
For maxima or minima put $f^{\prime}(x)=0$.
$\therefore \cos x-\sin x=0 $
$\Rightarrow \sin x=\cos x $
$\Rightarrow \frac{\sin x}{\cos x}=1$
$\Rightarrow \tan x=1 $
$\Rightarrow x=\frac{\pi}{4}, \frac{5 \pi}{4}, \ldots$
Now, $f^{\prime \prime}(x)$ will be negative when $(\sin x+\cos x)$ is positive i.e., when $\sin x$ and $\cos x$ are both positive. Also, we know that $\sin x$ and $\cos x$ both are positive in the first quadrant.
Then, $f^{\prime \prime}(x)$ will be negative when $x \in\left(0, \frac{\pi}{2}\right)$.
Thus, we consider $x=\frac{\pi}{4}$.
$f^{\prime \prime}\left(\frac{\pi}{4}\right) =-\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right) $
$ =-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}< 0$
By second derivative test, $f$ will be maximum at $x=\frac{\pi}{4}$ and the maximum value of $f$ is
$f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$