Question
Answers

The greatest integer function is not differentiable at integral points. Give a reason. Differentiate $\sin \sqrt x $ with respect to $x$

Answer
VerifiedVerified
91.5k+ views
Hint: In this question we will first give a proof as to why the greatest integer function is not differentiable at integral points and then solve the differentiation part of the question.

Complete step-by-step solution:
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.

The required answer is $\dfrac{{\operatorname{Cos} (\sqrt x )}}{{2\sqrt x }}$.

Note: At a point, a function is differentiable when it is derivative at that point. So we defined that the tangent line of the points from the left in the slope is approaching the same value as the tangent of the points from the right in the slope.