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, $f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ or $\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$ provided that the limit exists. If $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ 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 $=\mathrm{RHD}=R f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ and
Left Hand Derivative $=\operatorname{LHD}=L f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$
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 $x=a$

$
\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim _{h \rightarrow 0} \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 $A B=\frac{f(a+h)-f(a)}{(a+h)-a}$

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

$
\lim _{h \rightarrow 0}(\text { slope of chord } A B)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}
$
Right-hand derivative

$
=R\left[f^{\prime}(a)\right]=\lim _{h \rightarrow 0}(\text { slope of chord } A B)
$
As $h \rightarrow 0, B \rightarrow A$ on a curve, $a+h \rightarrow a$ on $x$-axis and $f(a+h) \rightarrow f(a)$ 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 $A C=\frac{f(a-h)-f(a)}{(a-h)-a}$
Now apply $\lim \mathrm{h} \rightarrow 0$ on both sides to get :

$
\lim _{h \rightarrow 0}(\text { slope of chord } A C)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}
$
Left-hand derivative

$
=L\left[f^{\prime}(a)\right]=\lim _{h \rightarrow 0}(\text { slope of chord } A C)
$
As $h \rightarrow 0, C \rightarrow A$ on the curve, $a-h \rightarrow a$ on the $x$-axis and $f(a-h) \rightarrow f(a)$ 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 $L\left[f^{\prime}(a)\right]=R\left[f^{\prime}(a)\right]$
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, $\quad \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$ exists finitely.
Let,

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

$
\begin{aligned}
& =\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \lim _{x \rightarrow a}(x-a) \\
& =f^{\prime}(\mathrm{a}) \times 0 \\
& =0
\end{aligned}
$
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