Differentiability of Composite Function

Properties of differentiable functions

1. Absolute functions are always continuous throughout but not differentiable at their critical point.
2. Properties of differentiable functions

At every corner point $f(x)$ is continuous but not differentiable. ex: $|x-a|$ is continuous but not differentiable at $\mathrm{x}=\mathrm{a}$ for $\mathrm{a}>0$

Differentiability in an Interval

A) A function $f(x)$ is differentiable in an open in interval ( $a, b$ ) if it is differentiable at every point on the open interval $(a, b)$.

B) A function $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is said to be differentiable in the closed interval [a, b].

1. If $f(x)$ is differentiable at every point on the open interval (a,
b). And,
2. It is differentiable from the right at " $a$ " and the left at " $b$ ".

(In other words, $\lim\limits_{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a}$ and $\lim\limits_{x \rightarrow b^{-}} \frac{f(x)-f(b)}{x-b}$ both exists), then $f(x)$ is said to be differentiable in $[a, b]$

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, $\quad \lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$ exists finitely.
Let,

$
f^{\prime}(a)=\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \qquad. . . (i)
$
In order to prove that $f(x)$ is continuous at $x=$ a it is enough to show that $\lim\limits_{x \rightarrow a} f(x)=f(a)$
or we have to show that $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}}[\mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{a})]=0$
Now,

$
\begin{aligned}
\lim\limits_{x \rightarrow a}[f(x)-f(a)] & =\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}(x-a) \\
& =\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \lim\limits_{x \rightarrow a}(x-a) \\
& =f^{\prime}(a) \times 0 \\
& =0
\end{aligned}
$
Therefore, $f(x)$ is continuous at a.

Let a function $f(x)$ is differentiable at $x=a$

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 $=1$

As we see in the graph, at $x=0$, it has a sharp edge. If the graph of a function has a sharp turn at some point $x=a$, then the function is not differentiable at $\mathrm{x}=\mathrm{a}$.

Note:

1. If a function is differentiable, then it must be continuous at that point
2. If a function is continuous, then it may or may not be differentiable at that point
3. If a function is not continuous, then it is definitely not differentiable at that point
4. If a function is not differentiable, then it may or may not be continuous at that point

Theorems of Differentiability

Theorem 1

If $f(x)$ and $g(x)$ are both differentiable functions at $x=a$, then the following functions are also differentiable at $\mathrm{x}=\mathrm{a}$.
(i) $\mathrm{f}(\mathrm{x}) \pm \mathrm{g}(\mathrm{x})$
(ii) $\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})$
(iii) $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}$, provided $\mathrm{g}(\mathrm{a}) \neq 0$

Theorem 2

If $f(x)$ is differentiable at $x=a$ and $g(x)$ is not differentiable at $x=a$, then $f(x) \pm g(x)$ will not be differentiable at $x=a$.

For example, $\cos (x)+|x|$ is not differentiable at $x=0$, as $\cos (x)$ is differentiable at $x=0$, but $|x|$ is not differentiable at $x=0$.

In other cases, $f(x) \cdot g(x)$ and $f(x) / g(x)$ may or may not be differentiable at $x$ = a, and hence should be checked using LHD-RHD, continuity or graph.

For example, if $f(x)=0$ (differentiable at $x=0$ ) and $g(x)=|x|$ (nondifferentiable at $x=0$ ). Their product is $f(x) \cdot g(x)=0$ which is differentiable. But if $f(x)=2, g(x)=|x|$, then $f(x) \cdot g(x)=2|x|$ is non-differentiable at $x=0$.

Theorem 3

If $f(x)$ and $g(x)$ both are nondifferentiable functions at $x=a$, then the function obtained by the algebraic operation of $f(x)$ and $g(x)$ may or may not be differentiable at $x=a$. Hence they should be checked.

For example, Let $f(x)=|x|$, not differentiable at $x=0$, and $g(x)=-|x|$ which is also not differentiable at $x=0$. Their sum $=0$ is differentiable and the difference $=2|\mathrm{x}|$ is not differentiable. So there is no definite rule.

Theorem 4

Differentiation of a continuous function may or may not be continuous.

Recommended Video Based on Differentiability of Composite Functions

Solved Examples Based On Differentiability of Composite Functions

Example 1:

Let $\mathrm{S}=\left\{\mathrm{t} \in \mathrm{R}: \mathrm{f}(x)=|x-\pi| \cdot\left(e^{|x|}-1\right) \sin |x|\right)$ is not differentiable at t$\}$. Then the set $S$ is equal to:
[JEE Main 2018]
1) 0
2) $\varnothing$
3) $\{0\}$
4) $\{\pi\}$

Solution

As we have learned
Properties of differentiable functions -
At every corner point $f(x)$ is continuous but not differentiable.
ex: $|\mathrm{x}-\mathrm{a}|$ is continuous but not differentiable at $\mathrm{x}=\mathrm{a}$ for $\mathrm{a}>0$
- wherein

We have to check Differntiability of $S$ at $x=0, \pi$
$
\begin{aligned}
& \text { at } x=\pi \\
& f(x)=|x-\pi|\left(e^{|x|}-1\right) \sin |x| \\
& = \begin{cases}(x-\pi)\left(e^x-1\right) \sin x & x>\pi \\
(x-\pi)\left(e^x-1\right) \sin x & x<\pi\end{cases} \\
& f^{\prime}(h+\pi)=(x-\pi)\left(e^x-1\right) \cos x+(x-p i) \sin x \cdot e^x+\left(e^x-1\right) \sin x \cdot 1 \\
& \text { at } x=\pi \quad:=0+0+0=0 \\
& \text { similalrly, } f^{\prime}(\pi-h)=0 \\
& \text { hence at } x=\pi f(x) \text { differentiable } \\
& \text { at } x=0 \\
& f(x)=|x-\pi|\left(e^{|x|}-1\right) \sin |x| \\
& = \begin{cases}-(x-\pi)\left(e^x-1\right) \sin x & x<0 \\
+(x-\pi)\left(e^x-1\right) \sin x & x>0\end{cases} \\
& f^{\prime}(0+h)=-\left[(x-\pi)\left(e^x-1\right) \cos x+\left(e^x-1\right) \sin x+(x-\pi) e^x \sin x\right] \\
& x=0 \quad ;=0+0+0 \\
& \text { similalry } f^{\prime}(0+h)=0 \\
& f(x) \text { is df ferentiable at } x=0
\end{aligned}
$

Example 2: Let $f(x)=15-|x-10| ; x \in \mathbf{R}$. Then the set of all values of $x$, at which the function, $g(x)=f(f(x))$ is not differentiable, is :
[JEE Main 2019]
1) $\{5,10,15\}$
2) $\{10,15\}$
3) $\{5,10,15,20\}$
4) $\{10\}$

Solution

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

$
\begin{aligned}
f(x) & =15-|x-10| \\
g(x) & =f(f(x))=f(15-|x-10|) \\
& =15-|5-| x-10|| \\
& =\left\{\begin{array}{cc}
15-|x-5| & x<10 \\
15-|15-x| & 10<x
\end{array}\right. \\
& =\left\{\begin{array}{cc}
10+x & x<5 \\
20-x & 5<x 10 \\
x & 10<x<15 \\
30-x & 15<x
\end{array}\right.
\end{aligned}
$

$f(x)$ is not differentiable at

$
x=5,10,15
$

Example 3: Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function defined as

$
f(x)=\left\{\begin{array}{ccc}
3\left(1-\frac{|x|}{2}\right) & \text { if } & |x| \leq 2 \\
0 & \text { if } & |x|>2_{\text {Let }}
\end{array}\right.
$

$g: \mathbf{R} \rightarrow \mathbf{R}$ be given by $g(x)=f(x+2)-f(x-2)_{\text {If }}$ n and m denote the number of points in $\mathbf{R}$ where $g$ is not continuous and not differentiable, respectively, then $\mathrm{n}+\mathrm{m}$ is equal to $\qquad$ [JEE Main 2021]
1) 4
2) 3
3) 2
4) 0

Solution
$
\begin{aligned}
& f(x+2)=\left\{\begin{array}{cl}
3\left(1-\frac{|x+2|}{2}\right) & \text { if }|x+2| \leq 2 \\
0 & \text { if }|x+2|>2
\end{array}\right. \\
& =\left\{\begin{array}{c}
3\left(1-\frac{|x+2|}{2}\right),-4 \leq x \leq 0 \\
0
\end{array}\right. \\
& \text { if } x>0 \text { or } x<-4 \\
& f(x-2)=\left\{\begin{array}{cl}
3\left(1-\frac{|x-2|}{2}\right) & \text { if }|x-2| \leq 2 \\
0 & \text { if }|x-2|>2
\end{array}\right. \\
& = \begin{cases}3\left(1-\frac{|x-2|}{2}\right) & \text { if } 0 \leq x \leq 4 \\
0 & \text { if } x<0 \text { or } x>4\end{cases} \\
& g(x)=f(x+2)+f(x-2) \\
& =\left\{\begin{array}{cl}
0 & x<-4 \text { or } x>4 \\
3\left(1-\frac{|x+2|}{2}\right) & ,-4 \leqslant x \leqslant 0 \\
3\left(1-\frac{|x-2|}{2}\right) & \text { if } \quad 0<x \leqslant 4
\end{array}\right.
\end{aligned}
$

Clearly $n=0, m=4$.

$
n+m=4 \text {. }
$
Hence, the answer is (4).

Example 4: Let

$
\mathrm{f}(\mathrm{x})= \begin{cases}\left|4 x^2-8 x+5\right|, & \text { if } 8 x^2-6 x+1 \geqslant 0 \\ {\left[4 x^2-8 x+5\right],} & \text { if } 8 x^2-6 x+1<0\end{cases}
$

where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the numbers of points in $\mathbf{R}$ where $f$ is not differentialble is $\qquad$
[JEE Main 2022]
1) 3
2) 2
3) 1
4) 0

Solution

Hence answer is 3

Example 5 : Let $f$ be a twice differentiable function on ( 1,6 . If $(2)=8, f^{\prime}(2)=5, f^{\prime}(x) \geq 1$ and $f^{\prime \prime}(x) \geq 4$, for all $x \in(1,6)_{\text {then: }}$
[JEE Main 2020]
1) $f(5)+f^{\prime}(5) \leq 26$
2) $f(5)+f^{\prime}(5) \geq 28$
3) $f(5)+f^{\prime \prime}(5) \leq 20$
4) $f(5) \leq 10$

Solution
$
\begin{aligned}
& \mathrm{f}(2)=8, \mathrm{f}^{\prime}(2)=5, \mathrm{f}^{\prime}(\mathrm{x}) \geq 1, \mathrm{f}^{\prime \prime}(\mathrm{x}) \geq 4, \forall \mathrm{x} \in(1,6) \\
& f^{\prime \prime}(x)=\frac{f^{\prime}(5)-f^{\prime}(2)}{5-2} \geq 4 \Rightarrow f^{\prime}(5) \geq 17 \\
& f^{\prime}(x)=\frac{f(5)-f(2)}{5-2} \geq 1 \Rightarrow f(5) \geq 11 \\
& f^{\prime}(5)+f(5) \geq 28
\end{aligned}
$
Hence, the answer is the option 2.