Separate the interval $\left[ 0,\dfrac{\pi }{2} \right]$ into subintervals in which the function $f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x$ is strictly increasing or strictly decreasing.

Hint: We will find the first derivative of the given function with respect to $x.$ Then we will find the subintervals of the given interval. Then we use the first derivative test for increasing and decreasing functions to find the intervals in which the given function is strictly increasing or strictly decreasing.

Complete step-by-step solution:
Consider the given function $f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x.$
We know that the derivative of ${{\sin }^{4}}x$ with respect to $x$ is $4{{\sin }^{3}}x\cos x$ and the derivative of ${{\cos }^{4}}x$ with respect to $x$ is $-4{{\cos }^{3}}x\sin x.$
Now we are going to find the first derivative of the whole function given here to get,
$\Rightarrow {f}'\left( x \right)=4{{\sin }^{3}}x\cos x-4{{\cos }^{3}}x\sin x$
Now let us take $-4\sin x\cos x$ out so that we get the function as,
$\Rightarrow {f}'\left( x \right)=-4\sin x\cos x\left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)$
We know that ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ and ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}.$
Using these identities, we can compute $\left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)$ as
$\Rightarrow \left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)=-\dfrac{1-\cos 2x}{2}+\dfrac{1+\cos 2x}{2}$
That can be written as
$\Rightarrow \left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)=\dfrac{-1+\cos 2x}{2}+\dfrac{1+\cos 2x}{2}$
Since the denominator of both the fractions are same, we will get
$\Rightarrow \left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)=\dfrac{-1+\cos 2x+1+\cos 2x}{2}=\dfrac{2\cos 2x}{2}$
The above equation is obtained because $-1+1=0.$
Now we will eliminate $2$ from the numerator and the denominator to get
$\Rightarrow \left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)=\cos 2x.$
Thus, the given function becomes
$\Rightarrow {f}'\left( x \right)=-4\sin x\cos x\cos 2x.$
Also, we know the identity given by $2\sin x\cos x=\sin 2x.$
We are given with $-4\sin x\cos x.$
This can be rewritten as $-4\sin x\cos x=-2\left( 2\sin x\cos x \right)=-2\sin 2x.$
Therefore, the given function becomes
$\Rightarrow {f}'\left( x \right)=-2\sin 2x\cos 2x.$
We again use the identity $2\sin x\cos x=\sin 2x.$
In this case, $x$ is replaced with $2x.$
So, we will get $-2\sin 2x\cos 2x=-\sin \left( 2\cdot 2x \right)=-\sin 4x.$
Now, we found the first derivative of the given function as ${f}'\left( x \right)=-\sin 4x.$
To find the subintervals of the given interval in which the given function exists, we will equate the first derivative of the function to zero.
$\Rightarrow {f}'\left( x \right)=0.$
That is, $-\sin 4x=0.$
$\Rightarrow \sin 4x=0.$
This implies $4x=0,\pi ,2\pi ,...$
So, the values of $x=0,\dfrac{\pi }{4},\dfrac{\pi }{2}.$
So, the subintervals are $\left[ 0,\dfrac{\pi }{4} \right]$ and $\left[ \dfrac{\pi }{4},\dfrac{\pi }{2} \right].$
Now we get, ${f}'\left( x \right)=-\sin 4x<0$ on the interval $\left[ 0,\dfrac{\pi }{4} \right]$ and ${f}'\left( x \right)=-\sin 4x>0$ on the interval $\left[ \dfrac{\pi }{4},\dfrac{\pi }{2} \right].$
Hence, by the first derivative test for increasing and decreasing functions, the function $f\left( x \right)$ is strictly decreasing on the interval $\left[ 0,\dfrac{\pi }{4} \right]$ and strictly increasing on the interval $\left[ \dfrac{\pi }{4},\dfrac{\pi }{2} \right].$

Note: The first derivative test for increasing and decreasing functions is given below:
A function $f\left( x \right)$ is decreasing on an interval $\left[ a,b \right],$ if the first derivative ${f}'\left( x \right)<0$ on the interval $\left[ a,b \right].$
A function $f\left( x \right)$ is increasing on an interval $\left[ a,b \right],$ if the first derivative ${f}'\left( x \right)>0$ on the interval $\left[ a,b \right].$