Differentiability of Composite Function

Definition of Composite Function

A composite function is formed when one function is applied to the result of another. In simple terms, if we have two functions $f(x)$ and $g(x)$, then their composite is written as $f(g(x))$. This means we first apply $g$ to $x$ and then apply $f$ to the result of $g(x)$. Understanding composite functions helps in analyzing how multiple transformations affect a single variable step-by-step.

The notation $f(g(x))$ represents the composition of functions, where the output of $g(x)$ becomes the input for $f(x)$. For example, if $f(x) = x^2$ and $g(x) = x + 3$, then $f(g(x)) = (x + 3)^2$. This concept is essential in calculus because it often appears in problems involving the chain rule of differentiation and the composition of continuous or differentiable functions.

How Function Composition Works with Examples

To better understand how composition works, consider an example:
Let $f(x) = 2x + 1$ and $g(x) = x^2$. Then,
$f(g(x)) = 2x^2 + 1$ and $g(f(x)) = (2x + 1)^2 = 4x^2 + 4x + 1$.
Notice that $f(g(x))$ and $g(f(x))$ are not the same — this shows that function composition is not commutative. The order in which you apply functions matters, which is crucial when studying differentiability of composite functions.

Importance of Continuity and Differentiability in Composite Functions

For a composite function $f(g(x))$ to be differentiable at a point $x = a$, two conditions must be met:

1. The inner function $g(x)$ must be differentiable at $x = a$.

2. The outer function $f(x)$ must be differentiable at $g(a)$.

If both functions are continuous and differentiable in their respective domains, the composite $f(g(x))$ will also be differentiable. This relationship is foundational to the chain rule, which expresses the derivative of $f(g(x))$ as $f'(g(x)) \cdot g'(x)$.

Properties of Differentiable Functions

Differentiable functions play a vital role in limits, differentiation. Understanding their properties helps in determining where a function can be smoothly graphed without sharp edges or breaks. Below are the key properties and theorems related to differentiability and continuity.

1. Continuity and Differentiability of Absolute Functions

Absolute value functions are an important example to understand the relationship between continuity and differentiability.

• Absolute functions are always continuous throughout their domain but not differentiable at their critical point.

• At every corner point, a function $f(x)$ is continuous but not differentiable.

For example:
The function $f(x) = |x - a|$ is continuous but not differentiable at $x = a$, where $a > 0$.

This happens because at the point $x = a$, the left-hand derivative (LHD) and right-hand derivative (RHD) are not equal, creating a sharp corner in the graph.

Differentiability in an Interval

Differentiability is not limited to individual points—it can also be studied over intervals. Depending on the interval type (open or closed), the definition slightly varies.

A) Differentiability in an Open Interval

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

B) Differentiability in a Closed Interval

A function $y = f(x)$ is said to be differentiable in the closed interval $[a, b]$ if:

1. $f(x)$ is differentiable at every point on the open interval $(a, b)$, and

2. $f(x)$ is differentiable from the right at $a$ and from the left at $b$.

In mathematical terms:

$\lim\limits_{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a}$ and $\lim\limits_{x \rightarrow b^{-}} \frac{f(x)-f(b)}{x-b}$

both exist.

If these conditions hold, then $f(x)$ is said to be differentiable in $[a, b]$.

Chain Rule – Definition, Formula, and Examples

The Chain Rule is a fundamental concept in calculus that helps you find the derivative of a composite function, where one function is nested inside another. If you have a composite function of the form $y = f(g(x))$, then the Chain Rule states that the derivative of $y$ with respect to $x$ is obtained by differentiating the outer function first and then multiplying it by the derivative of the inner function.

Mathematically, the Chain Rule is expressed as:
$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

This means you first differentiate the outer function $f(u)$ with respect to its argument $u = g(x)$, and then multiply by the derivative of $g(x)$ with respect to $x$.

For example, if $y = \sin(3x^2)$, then
$\frac{dy}{dx} = \cos(3x^2) \times (6x) = 6x \cos(3x^2)$

The Chain Rule is essential in problems involving nested functions, parametric equations, and implicit differentiation. It also plays a key role in ensuring the differentiability of composite functions — since both the inner and outer functions must themselves be differentiable for $f(g(x))$ to be differentiable.

Theorem: Differentiability Implies Continuity

If a function $f(x)$ is differentiable at every point in an interval, then it must be continuous in that interval.
However, the converse (that every continuous function is differentiable) may or may not be true.

Proof

Let a function $f(x)$ be differentiable at $x = a$.
Then, by definition,

$f'(a) = \lim\limits_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} \quad ...(i)$

To prove that $f(x)$ is continuous at $x = a$, we must show that:

$\lim\limits_{x \rightarrow a} f(x) = f(a)$

or equivalently,

$\lim\limits_{x \rightarrow a} [f(x) - f(a)] = 0$

Now,

$\begin{aligned}
\lim\limits_{x \rightarrow a}[f(x)-f(a)] &= \lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} (x-a) \\
&= \left( \lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \right) \left( \lim\limits_{x \rightarrow a}(x-a) \right) \\
&= f'(a) \times 0 \\
&= 0
\end{aligned}$

Therefore, $f(x)$ is continuous at $x = a$.

Converse of the Theorem

The converse of the above theorem is not true.
That is, if a function is continuous at a point, it may or may not be differentiable at that point.

Example: The Modulus Function

Consider the function $f(x) = |x|$.
The modulus function is continuous at $x = 0$ but not differentiable at that point.

At $x = 0$:
LHD $= -1$ and RHD $= 1$

Since LHD $\ne$ RHD, the derivative does not exist at $x = 0$.

Graphically, at $x = 0$, the graph of $|x|$ has a sharp corner (cusp). Whenever a graph has such a sharp turn or non-smooth transition, the function becomes non-differentiable at that point.

Every differentiable function is guaranteed to be continuous, but not every continuous function is differentiable.
Functions like $|x|$, $|x - a|$, or piecewise-defined functions often serve as prime examples to understand non-differentiability at sharp points.

Key Notes on Continuity and Differentiability

Before studying the theorems of differentiability, it’s essential to understand how continuity and differentiability relate to one another. These properties often overlap but are not always interchangeable.

Important Points to Remember

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.

These points summarize the logical relationship between continuity and differentiability, helping you quickly identify which functions are smooth (differentiable) and which may have corners or breaks (non-differentiable points).

Theorems of Differentiability

Differentiability theorems form the foundation for operations on differentiable functions. They define how combinations, products, or quotients of differentiable and non-differentiable functions behave at specific points.

Theorem 1: Operations on Differentiable Functions

If $f(x)$ and $g(x)$ are both differentiable functions at $x = a$, then the following functions are also differentiable at $x = a$:

1. $f(x) \pm g(x)$

2. $f(x) \cdot g(x)$

3. $\frac{f(x)}{g(x)}$, provided that $g(a) \ne 0$

This means that addition, subtraction, multiplication, and division of two differentiable functions (where defined) yield a function that is also differentiable.

Theorem 2: Sum of a Differentiable and Non-Differentiable Function

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 $x = a$.

Example:
Consider $f(x) = \cos x$ and $g(x) = |x|$.
Here, $\cos(x)$ is differentiable at $x = 0$, but $|x|$ is not differentiable at $x = 0$.

Therefore, the function $\cos(x) + |x|$ is not differentiable at $x = 0$.

However, in the case of multiplication or division, differentiability is not guaranteed and must be checked separately using Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) or by verifying continuity and graph behavior.

Example:
Let $f(x) = 0$ (which is differentiable at $x = 0$) and $g(x) = |x|$ (non-differentiable at $x = 0$).
Then,
$f(x) \cdot g(x) = 0$ which is differentiable everywhere.

But if $f(x) = 2$ and $g(x) = |x|$, then $f(x) \cdot g(x) = 2|x|$, which is not differentiable at $x = 0$.

Hence, product or quotient forms must always be checked explicitly for differentiability.

Theorem 3: Operations on Two Non-Differentiable Functions

If both $f(x)$ and $g(x)$ are non-differentiable functions at $x = a$, the resulting function obtained through algebraic operations may or may not be differentiable at $x = a$.

Therefore, it is essential to verify differentiability through limits, continuity, or by checking LHD and RHD.

Example:
Let $f(x) = |x|$ (not differentiable at $x = 0$) and $g(x) = -|x|$ (also not differentiable at $x = 0$).

Then,

• Their sum $f(x) + g(x) = 0$ is differentiable.

• Their difference $f(x) - g(x) = 2|x|$ is not differentiable at $x = 0$.

This shows there is no definite rule, and each case must be checked individually.

Theorem 4: Differentiability Does Not Imply Continuity of Derivative

Even if a function is differentiable, the derivative of that function may or may not be continuous.

This means that while the function itself is smooth, its rate of change (derivative) might have discontinuities or jumps at certain points.

For example, consider piecewise functions that are differentiable everywhere but have a derivative that changes abruptly at certain points—these highlight that differentiation and continuity of the derivative are distinct properties.

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.

List of topics related to Differentiability of Composite Functions

This section lists important subtopics and formulas linked to the differentiability of composite functions, helping you understand how function composition and derivative rules work together in calculus.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Derivative as Rate Measure: Definition, Formula, Examples

Continuity of Composite Function

Continuity And Differentiability

NCERT Resources

This part compiles all NCERT study materials related to this chapter - including notes, textbook solutions, and exemplar problems - to make your preparation complete and exam-ready.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on Differentiability of Composite Functions

Test your understanding of composite function differentiability through a curated set of practice questions and MCQs aligned with the latest exam trends.

Differentiability Of Composite Function- Practice Question MCQ

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