Differentiability and Existence of Derivative - Practice Questions & MCQ

Differentiability and Existence of Derivative

The derivative of a function $\mathrm{f}(\mathrm{x})$ at point a is defined by $f^{\prime }(x)$ at $\mathrm{x}=$ a i.e, $f^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$ or $\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$ provided that the limit exists. If $\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$ does not exists, we say that the function $\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=\mathrm{a}$.

Or, we can say that a function $f(x)$ is differentiable at a point 'a' in its domain if the limit of the function $f^{\prime }(x)$ exists at $x=a$.
i.e.

Right Hand Derivative $=\mathrm{RHD}=R{f}^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$ and
Left Hand Derivative $=\mathrm{LHD}=L{f}^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a-h)-f(a)}{-h}$
are finite and equal.
(Both the left and right-hand derivatives are finite and equal.)
$R{f}^{\prime }(a)=L{f}^{\prime }(a)$ is the condition for differentiability at $x=a$

$\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}=\underset{h\to 0}{lim}\frac{f(a-h)-f(a)}{-h}$

Geometrical meaning of Right-hand derivative

Let $A(a,f(a))$ and $B(a+h,f(a+h))$ be two points very near to each other on the curve $y=f(x)$. Using the slope of a line formula, we get
Slope of $AB=\frac{f(a+h)-f(a)}{(a+h)-a}$

Now apply $lim\mathrm{h}\to 0$ on both sides to get :

$\underset{h\to 0}{lim}(\text{ slope of chord }AB)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$
Right-hand derivative

$=R[f^{\prime }(a)]=\underset{h\to 0}{lim}(\text{ slope of chord }AB)$
As $h\to 0,B\to A$ on a curve, $a+h\to a$ on $x$-axis and $f(a+h)\to f(a)$ on y-axis.
When h is infinitely small, chord AB almost becomes tangent drawn at A towards the right.
Hence, the right-hand derivative's geometrical significance is that it represents the slope of the tangent drawn at A towards the right.

Geometrical meaning of Left-hand derivative

Let $A(a,f(a))$ and $C(a-h,f(a-h))$ be two points very near to each other on the curve $y=f(x)$. Using the slope of a line formula, we get
Slope of $AC=\frac{f(a-h)-f(a)}{(a-h)-a}$
Now apply $lim\mathrm{h}\to 0$ on both sides to get :

$\underset{h\to 0}{lim}(\text{ slope of chord }AC)=\underset{h\to 0}{lim}\frac{f(a-h)-f(a)}{-h}$
Left-hand derivative

$=L[f^{\prime }(a)]=\underset{h\to 0}{lim}(\text{ slope of chord }AC)$
As $h\to 0,C\to A$ on the curve, $a-h\to a$ on the $x$-axis and $f(a-h)\to f(a)$ on the $y$-axis.
When h is infinitely small, chord AC almost becomes tangent drawn at A towards the left.
Hence the geometrical significance of the left-hand derivative is that it represents the slope of the tangent drawn at A towards the left.

Geometrical meaning of the existence of derivative
We know that the derivative of a function exists at $x=a$, if $L[f^{\prime }(a)]=R[f^{\prime }(a)]$
1. The slope of the tangent drawn at $A$ towards left = slope of the tangent drawn at $A$ towards the right
2. The same tangent line towards left and right: meaning a unique tangent at the point
3. Smooth curve around $x=a$

Differentiability and Continuity

Theorem

If a function $f(x)$ is differentiable at every point in an interval, then it must be continuous in that interval. But the converse may or may not be true.
Proof
Let a function $f(x)$ is differentiable at $x=a$
then, $\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$ exists finitely.
Let,

$f^{\prime }(a)=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$
To prove that $f(x)$ is continuous at $x=a$ it is enough to
show that $\underset{x\to a}{lim}f(x)=f(a)$
or we have to show that $\underset{x\to a}{lim}[f(x)-f(a)]=0$
Now, $\underset{x\to a}{lim}[f(x)-f(a)]=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}(x-a)$

$\begin{array}{rl}& =\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}\underset{x\to a}{lim}(x-a)\\ & =f^{\prime }(\mathrm{a})\times 0\\ & =0\end{array}$
Therefore, $f(x)$ is continuous at a.

Converse

The converse of the above theorem is NOT true. i.e. if a function is continuous at a point then it may or may not be differentiable at that point.
For example,
Consider the function, $f(x)=|x|$, the modulus function is continuous at $x=0$ but it is not differentiable at $x=0$ as LHD = -1 and RHD =
As we see in the graph, at $\mathrm{x}=0$, it has a sharp edge. If the graph of a function has a sharp turn at some point $\mathrm{x}=\mathrm{a}$, then the function is not differentiable at $\mathrm{x}=\mathrm{a}$.

Note:

If a function is differentiable, then it must be continuous at that point

If a function is continuous, then it may or may not be differentiable at that point

If a function is not continuous, then it is not differentiable at that point

If a function is not differentiable, then it may or may not be continuous at that point