Find the interval in which function f(x) = sinx+cosx is increasing

The correct answer is [0, π/4) and (5π/4, 2π].

To find the interval in which the function is increasing, we need to analyze its derivative. A function $f\left(x\right)$ is strictly increasing in an interval where its first derivative is positive, i.e., $f^{\prime }\left(x\right)>0$.

Let the given function be:
$f\left(x\right)=sinx+cosx$

Differentiating $f\left(x\right)$ with respect to $x$ gives:
$f^{\prime }\left(x\right)=\frac{d}{dx}(sinx+cosx)=cosx-sinx$

For the function to be increasing, we require:
$f^{\prime }\left(x\right)>0$
$\Rightarrow cosx-sinx>0$
$\Rightarrow cosx>sinx$

Assuming the standard domain of $[0,2\pi ]$ based on the options, let us compare the values of $cosx$ and $sinx$:
1. At $x=0$, $cos0=1$ and $sin0=0$, so $cosx>sinx$ holds true.
2. In the interval $[0,\pi /4)$, $cosx>sinx$ because the cosine value starts at 1 and decreases to $1/\sqrt{2}$, whereas the sine value starts at 0 and increases to $1/\sqrt{2}$.
3. At $x=\pi /4$, both are equal: $cos(\pi /4)=sin(\pi /4)=1/\sqrt{2}$.
4. Between $\pi /4$ and $5\pi /4$, $sinx$ is greater than $cosx$.
5. At $x=5\pi /4$, both are equal again: $cos(5\pi /4)=sin(5\pi /4)=-1/\sqrt{2}$.
6. For $x$ in the interval $(5\pi /4,2\pi ]$, the cosine function is larger than the sine function (in the fourth quadrant, cosine is positive while sine is negative, and in the latter part of the third quadrant, cosine is less negative than sine).

Thus, $cosx>sinx$ is satisfied in the intervals:
$[0,\frac{\pi }{4})\cup (\frac{5\pi }{4},2\pi ]$

Therefore, the function $f\left(x\right)$ is increasing in the interval [0, π/4) and (5π/4, 2π].