Check whether the given function f(x) is differentiable at x=1 or not.
$f\left( x \right)=\sqrt{1-{{x}^{2}}}$

Hint: First, before proceeding for this, we must know the formula for the differentiability given by the first principle of derivative rule as ${f}'\left( x \right)=\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}$. Then, we also know the condition that the above derivative function should give the value of the derivative as positive or negative derivative to get it as differentiable. Then, by using the condition stated and substituting the value of a as 1 and solving the limits, we get the desired result.

Complete step by step answer:
In this question, we are supposed to check whether the given function $f\left( x \right)=\sqrt{1-{{x}^{2}}}$is differentiable at x=1 or not.
So, before proceeding for this, we must know the formula for the differentiability given by the first principle of derivative rule as:
${f}'\left( x \right)=\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}$
Now, we also know the condition that the above derivative function should give the value of the derivative as positive or negative derivative to get it as differentiable.
Now, by using it to test the differentiability at x=1, we will substitute the value of a as 1.
Then, we need to find the function f(a) by replacing x with a in f(x) as:
$f\left( a \right)=\sqrt{1-{{a}^{2}}}$
Then, by substituting the value of f(a) and f(x) in the first principle derivative formula, we get:
${f}'\left( x \right)=\displaystyle \lim_{x \to a}\dfrac{\sqrt{1-{{x}^{2}}}-\sqrt{1-{{a}^{2}}}}{x-a}$
Now, firstly we will check whether after substituting the value of point it does not approach infinity.
So, by substituting the value of a as 1, we get:
$\begin{align}
  & {f}'\left( x \right)=\displaystyle \lim_{x \to 1}\dfrac{\sqrt{1-{{x}^{2}}}-\sqrt{1-{{1}^{2}}}}{x-1} \\
 & \Rightarrow {f}'\left( x \right)=\displaystyle \lim_{x \to 1}\dfrac{\sqrt{1-{{x}^{2}}}-0}{x-1} \\
 & \Rightarrow {f}'\left( x \right)=\displaystyle \lim_{x \to 1}\dfrac{\sqrt{1-{{x}^{2}}}}{x-1} \\
\end{align}$
Now, by solving the limits of the above expression, we get:
$\begin{align}
  & {f}'\left( x \right)=\dfrac{\sqrt{1-{{1}^{2}}}}{1-1} \\
 & \Rightarrow {f}'\left( x \right)=\dfrac{\sqrt{1-1}}{0} \\
 & \Rightarrow {f}'\left( x \right)=\infty \\
\end{align}$
So, we can clearly see that derivative of the function at point x=1 gives the value infinity which means function $f\left( x \right)=\sqrt{1-{{x}^{2}}}$ is not differentiable at x=1.

Hence, $f\left( x \right)=\sqrt{1-{{x}^{2}}}$ is not differentiable at x=1.

Note: Now, to solve these types of the questions we need to know some of the basics of differentiability beforehand. As in the above question the first condition of differentiability is not satisfied which gives the function not differentiable. If it gets satisfied then we need to check LHD(left hand derivative) by substituting (x=a+h) and RHD(right hand derivative) by substituting (x=a-h) to be equal given by the formula:
$\displaystyle \lim_{x \to 0}\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}=\displaystyle \lim_{x \to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$