Question

Find the maximum value of $\sin x+\cos x$ (using differentiation).

Answer 1

Let $f\left(x\right)=\sin x+\cos x$ ...(i)

Differentiating $f^{\prime }\left(x\right)=\cos x-\sin x$

$f^{\prime \prime }\left(x\right)=\sin x-\cos x$ ...(ii)

For maximum or minimum $f^{\prime }\left(x\right)=0$.

$\therefore \cos x-\sin x=0$

$\Rightarrow \sin x=\cos x$

$\Rightarrow \tan x=1$

$\Rightarrow x=\frac{\pi }{4}$.

Differentiating equation (2), $f^{\prime \prime }\left(x\right)=-\sin x-\cos x$

Putting $x=\frac{\pi }{4}$ in above equation $f^{\prime \prime }\left(\frac{\pi }{4}\right)=-\sin \frac{\pi }{4}-\cos \frac{\pi }{4}$

$=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$

$=\frac{-2}{\sqrt{2}}=-\sqrt{2}$.

$\because f^{\prime \prime }\left(\frac{\pi }{4}\right)$ is negative, hence at $x=\frac{\pi }{4}$ function had maximum value.

Now putting $x=\frac{\pi }{4}$ in equation (1)

$\therefore$Maximum value $f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}$

$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$.