Examining differentiability Using Graph of Function

Differentiability is an important concept in calculus. It is useful in understanding the rate of a change in the function. The existence of a derivative at a point implies that the function has a specific rate of change at that point. These concepts of differentiability have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

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In this article, we will cover the concept of the Condition of differentiability. This topic falls under the broader category of Calculus, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of nine questions have been asked on this topic in JEE Main from 2013 to 2023, one question in 2014, two questions in 2019, four in 2021, and two in 2023.

Condition for differentiability

A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise nondifferentiable

Suppose f is a real function and c is a point in its domain. The derivative of $f$ at c is defined by

$
\lim\limits_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}
$

provided this limit exists.
Derivative of f at c is denoted by $\mathrm{f}^{\prime}$ (c) or $\left.\frac{d}{d x}(f(x))\right|_c$
The function defined by

$
f^{\prime}(x)=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
$
The process of finding the derivative of a function is called differentiation. We also use the phrase differentiate $f(x)$ concerning $x$ to mean find $f$ '(x)Whenever we defined derivative, we had put a caution provided the limit exists. Now the natural question is; what if it doesn't? The question is quite pertinent and so is its answer. If $\lim\limits_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$

it does not exist, we say that $f$ is not differentiable at $c$. In other words, we say that a function f is differentiable at a point c in its domain if both $\lim\limits_{h \rightarrow 0^{-}} \frac{f(c+h)-f(c)}{h} \lim\limits_{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ are finite and equal.

Examining Differentiability Using Differentiation and Graph

1. Using Differentiation (only for continuous functions)

Some functions are defined piecewise, in such cases first we need to check if the function is continuous at the split point, and if it is continuous we need to differentiate each branch function and compare left-hand and right-hand derivative at the split point.

$
f(x)= \begin{cases}g_1(x), & x<a \\ g_2(x), & x \geq a\end{cases}
$
First check if $f(x)$ is continuous at $x=a$. If it is not continuous, then it cannot be differentiable. If it is continuous, then to check differentiability, find

$
f^{\prime}(x)= \begin{cases}\left(g_1(x)\right)^{\prime}, & x<a \\ \left(g_2(x)\right)^{\prime}, & x>a\end{cases}
$
Differentiability can be checked at $\mathrm{x}=\mathrm{a}$ by comparing

$
\lim\limits_{x \rightarrow a^{-}}\left(g_1(x)\right)^{\prime} \text { and } \lim\limits_{x \rightarrow a^{+}}\left(g_2(x)\right)^{\prime}
$

2. Differentiability using Graphs

A function $f(x)$ is not differentiable at $x=a$ if
1. The function is discontinuous at $x=a$
2. The graph of a function has a sharp turn at $x=a$
3. A function has a vertical tangent at $x=a$

Illustration 1

Check the differentiability of the following function.
1. $f(x)=\sin |x|$

Method 1

Using graphical transformation, we can draw its graph

Using the graph we can tell that at $x=0$, the graph has a sharp turn, so it is not differentiable at $x=0$.

Method 2

As $L H L=R H L=f(0)=0$, so the function is continuous at $x=0$
So we can use differentiation to check differentiability

$
\begin{aligned}
& \quad \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}
-\sin x, & x<0 \\
\sin x, & x \geq 0
\end{array}\right. \\
& \therefore \quad \quad \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc}
-\cos x, & x<0 \\
\cos x, & x>0
\end{array}\right. \\
& \therefore \quad \text { LHD }=\mathrm{f}^{\prime}\left(0^{-}\right)=-1 \text { and } \mathrm{RHD}=\mathrm{f}^{\prime}\left(0^{+}\right)=1
\end{aligned}
$
As these are not equal, so, $f(x)=\sin |x|$ is not differentiable at $x=0$

Illustration 2

$
f(x)=\|\log \mid x\|, x \text { not equal to } 0
$
Plot the graph of | log $|\mathrm{x}|$ | using graphical transformation

We can see that graph has a sharp turn at +1 and -1 so the function is not differentiable at these points.

Recommended Video Based on Condition of Differentiability

Solved Examples Based On Condition of Differentiability:

Example 1: Let $f, g: R \rightarrow R$ Be two functions defined by $f(x)=\left\{\begin{array}{cl}x \sin \left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x=0\end{array}\right.$ and $g(x)=x f(x)$: [JEE Main 2014]

Statement I :$f$ is a continuous function at $\mathrm{x}=0$.
Statement II : g is a differentiable function at $\mathrm{x}=0$.
1) Both statements I and II are false.
2) Both statements I and II are true.
3) Statement I is true, statement II is false.
4) Statement I is false, statement II is true.

Solution

As we learned in
Condition for differentiable -
A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x=x_0$ if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise non differentiable

$
\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x \sin \left(\frac{1}{x}\right) & x \neq 0 \\
0 & x=0
\end{array}\right. \\
& \text { and } g(x)=x f(x) \\
& \lim\limits_{x \rightarrow 0^{+} / 0^{-}} \quad x \sin \frac{1}{x}=0 \times \text { finite }=0
\end{aligned}
$
So $f(x)$ is continuous at $x=0$

$
\begin{aligned}
& g(x)=\left\{\begin{array}{cl}
x^2 \sin \frac{1}{x} & x \neq 0 \\
0 & x=0
\end{array}\right. \\
& \lim\limits_{h \rightarrow 0^{+}} \frac{h^2 \sin \frac{1}{h}-0}{h}=h \sin \frac{1}{h}=0 \\
& \lim\limits_{h \rightarrow 0^{-}} \frac{-h^2 \sin \frac{1}{h}-0}{-h}=h \sin \frac{1}{h}=0 \\
& \therefore g^{\prime}\left(0^{+}\right)=g^{\prime}\left(0^{-}\right)
\end{aligned}
$

So, $\mathrm{g}(\mathrm{x})$ is differentiable at $x=0$

Example 2: Let $f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^2-1, & 0 \leq x \leq 2\end{array}\right.$ and $g(x)=|f(x)|+f(|x|)$.Then, in the interval $(-2,2), \mathrm{g}$ is: [JEE Main 2019]
1) not differentiable at two points
2) differentiable at all points
3) not differentiable at one point
4) not continuous

Solution

Condition for differentiability -
A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x=x_{\circ}$ if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise non differentiable

Properties of differentiable functions -
At every corner point $f(x)$ is continuous but not differentiable.
ex: $|x-a|$ is continuous but not differentiable at $x=a$ for $a>0$
- wherein

$
y=f(x)
$

only one nondifferential point at $x=1$

Example 3: Let $f:(-1,1) \rightarrow R_{\text {be a function defined by }}$ $f(x)=\max \left\{-|x|,-\sqrt{1-x^2}\right\}$. If K be the set of all points at which $f$ is not differentiable, then K has exactly: [JEE Main 2019]
1) two elements
2) three elements
3) five elements
4) one element

Solution

Properties of differentiable functions -
At every corner point $f(x)$ is continuous but not differentiable.
ex: $|x-a|$ is continuous but not differentiable at $x=a$ for $a>0$
- wherein

Condition for differentiability -
A function $\mathrm{f}(\mathrm{x})$ is said to be differentiable at $x=x_{\circ}$ if $R f^{\prime}\left(x_{\circ}\right)$ and $L f^{\prime}\left(x_{\circ}\right)$ both exist and are equal otherwise non differentiable

$
\begin{aligned}
& f:(-1,1) \rightarrow R \\
& f(x)=\max \left\{-|x|, \sqrt{1-x^2}\right\}
\end{aligned}
$
Non-differentiable at 3 points in $(-1,1)$.

Example 4: Let $f:[0,3] \rightarrow \mathbf{R}_{\text {be defined by }}$ $f(x)=\min \{x-[x], 1+[x]-x\}$ where $[x]$ is the greatest integer less than or equal to x . Let P denote the set containing all $\mathrm{x} \in[0,3]_{\text {where } \mathrm{f}}$ is discontinuous, and Q denote the set containing all $x \in(0,3)$ where $f$ is not differentiable. Then the sum of number of elements in $P$ and $Q$ is equal to $\qquad$ [JEE Main 2021]
1) 5
2) 4
3) 2
4) 1

Solution

$f(x)=\min (\{x\}, 1-\{x\})$

No point of discontinuity $\Rightarrow n(P)=0$
5 points of non-differentiability $\Rightarrow n(Q)=5$

$
\Rightarrow n(P)+n(Q)=5
$

Example 5: Let $f:[0, \infty) \rightarrow[0,3]{\text {be a function }}$ defined by $f(\mathrm{x})= \begin{cases}\max \{\sin t: 0 \leq \mathrm{t} \leq \mathrm{x}\}, & 0 \leq \mathrm{x} \leq \pi \\ 2+\cos \mathrm{x}, & \mathrm{x}>\pi\end{cases}$ Then which of the following is true? [JEE Main 2021]

1) $f$ is continuous everywhere but not differentiable exactly at one point in $(0, \infty)$
2) $f$ is differentiable everywhere in $(0, \infty)$
3) $f$ is not continuous exactly at two points in $(0, \infty)$
4) $f$ is continuous everywhere but not differentiable exactly at two points in $(0, \infty)$

Solution

Is differentiable everywhere in $(0, \infty)$
The option (2) is correct.

Summary: The derivative is an important concept of the limits. Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point. With the help of differentiation, we can find the rate of change of one quantity for another. The concept of differentiation is the cornerstone on which the development of calculus rests.