Careers360 Logo
JEE Main 2025 - Exam Date, Eligibility, Registration, Preparation Tips, Previous Year Papers

Differentiability and Existence of Derivative - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Differentiability and Existence of Derivative is considered one the most difficult concept.

  • 202 Questions around this concept.

Solve by difficulty

Let f:R\rightarrow R   be defined by

f(x)=\left\{\begin{matrix} k-2x, & i\! fx\leq -1\\ 2x+3, & i\! fx> -1 \end{matrix}\right.

If f has a local minimum at x=-1, then a possible value of k is

Concepts Covered - 2

Differentiability and Existence of Derivative

Differentiability and Existence of Derivative

The derivative of a function f(x) at point a is defined by \\f'(x) \text{ at x =a i.e, } \;f'(a) = \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}\;\;\text{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 f(x) is not differentiable at x = a.

Or, we can say that a function f(x) is differentiable at a point ‘a’ in its domain if limit of the function f’(x) exists at x = a.

i.e.

\\\text { Right Hand Derivative = RHD}=R f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\\\text{and}\\\text { Left Hand Derivative = LHD}=L f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}

are finite and equal.

(Both the left-hand derivative and the right-hand derivative are finite and equal.)

 

\\R f^{\prime}(a)=L f^{\prime}(a) \text { 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

\text { Slope of } AB=\frac{f(a+h)-f(a)}{(a+h)-a}

Now apply lim h⇾0  on both sides to get :

\\\lim _{h \rightarrow 0}(\text { slope of chord } AB)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \\\\ {\text { Right hand derivative }} \\ {\quad \quad=R\left[f^{\prime}(a)\right]=\lim _{h \rightarrow 0}(\text { slope of chord } AB)}

As h⇾0, B ⇾ A on a curve, a + h ⇾ a on x-axis and f (a + h) ⇾ f (a) on y-axis.

When h is infinitely small, chord AB almost becomes tangent drawn at A towards the right.

Hence the geometrical significance of right-hand derivative 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

\text { Slope of } AC=\frac{f(a-h)-f(a)}{(a-h)-a}

Now apply lim h⇾0  on both sides to get :

\\\lim _{h \rightarrow 0}(\text { slope of chord } AC)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h} \\\\ {\text { Left hand derivative }} \\ {\quad \quad=L\left[f^{\prime}(a)\right]=\lim _{h \rightarrow 0}(\text { slope of chord } AC)}

As h⇾0, C ⇾ A on the curve, a - h ⇾ a on the x-axis and f (a - h) ⇾ f (a) on the y-axis.

When h is infinitely small, chord AC almost becomes tangent drawn at A towards 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’(a)] = R[f’(a)]

  1. 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

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

\\\mathrm{then,\;\;\;\;\;\;\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\;exists\;finitely.}\\\\\mathrm{Let,\;\;\;\;\;\;\;\;\;\;\;\;\;f^{\prime}(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(i)}\\\\\text{In order to prove that f(x) is continuous at x = a it is enough to}\\\\\mathrm{show \;that\;\;\lim_{x\rightarrow a}\;f(x)=f(a)}\\\\\mathrm{or\;we\;have\;to\;show\;that\;\;\lim_{x\rightarrow a}\;\left [f(x)-f(a) \right ]=0}\\\\\mathrm{Now,\;\;\;\;\;\;\lim_{x\rightarrow a}\;\left [f(x)-f(a) \right ]=\lim_{x\rightarrow a}\;\frac{f(x)-f(a)}{x-a}(x-a)}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim_{x\rightarrow a}\;\frac{f(x)-f(a)}{x-a}\lim_{x\rightarrow a}(x-a)}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=f'(a)\times 0}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=0}

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 = 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 x = 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

Study it with Videos

Differentiability and Existence of Derivative
Differentiability and Continuity

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Books

Reference Books

Differentiability and Existence of Derivative

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 4.17

Line : 33

Differentiability and Continuity

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 4.18

Line : 1

E-books & Sample Papers

Get Answer to all your questions

Back to top