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The Greatest Integer Function is denoted by \[y{\text{ }} = {\text{ }}\left[ x \right]\] which means it is less than or equal to $x$.

It rounds a real number to the nearest or the closest integer.

Greatest integer function isn't continuous at the integers level and any function which is discontinuous at the integer value, will be non−differentiable at that point.

As the value jumps at each integral value, therefore, it is discontinuous at each integral value.

At the Integral points, the Left-Hand of a function $ \ne $ Right-Hand Limit of the function.

Let’s consider an example, the greatest integer of 2.

We can write it as,

Left Hand Limit = \[\mathop {\lim }\limits_{h \to {0^ + }} f(2 - h) = 1\] (as h is a very small integer greatest integer to the left of $2 - h = 1$).

Also,

Right Hand Limit = \[\mathop {\lim }\limits_{h \to {0^ - }} f(2 - h) = 2\] (as h is a very small integer greatest integer to the right of\[2 - h = 2\]).

So, Left-Hand Limit $ \ne $ Right-Hand Limit

Hence Proved

Now we have to find out the differentiation of $\sin \sqrt x $ with respect to $x$

$ \Rightarrow \dfrac{{dy}}{{dx}}\operatorname{Sin} (\sqrt x )$

During derivative a chain rule is used in which there is a composite function such as $f(g(x))$ then its differentiation will be $f'g(x) \times g'(x)$.

Now we use the chain rule here, we get:

$ \Rightarrow \dfrac{{dy}}{{dx}}\operatorname{Sin} (\sqrt x ) \times \dfrac{{dy}}{{dx}}(\sqrt x )$

Here we know the general differentiation for $\dfrac{{dy}}{{dx}}\operatorname{Sin} (x) = \operatorname{Cos} (x)$ and $\dfrac{{dy}}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$

$ \Rightarrow \operatorname{Cos} (\sqrt x ) \times \dfrac{1}{{2\sqrt x }}$

Upon simplifying we get:

$ \Rightarrow \dfrac{{\operatorname{Cos} (\sqrt x )}}{{2\sqrt x }}$,

Hence we get the required answer.