$\text{f}(\text{x})=\sin\text{x}+\cos\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
Since, $\sin\text{x}$ and $\cos\text{x}$ are everywhere differentiable and continuous,
$\text{f}(\text{x})=\sin\text{x}+\cos\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big]$ and differentiable on $\Big(0,\frac{\pi}{2}\Big)$
Also,
$\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=1$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have show that there must exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\sin\text{x}+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sin\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\cos\text{x}-\sin\text{x}=0$
$\Rightarrow\tan\text{x}=1$
$\Rightarrow\text{x}=\frac{\pi}{4}$
Thus, $\text{c}=0\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
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$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$