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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

EASY

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

–1/2

–1

View Solution

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)]

Slope of the tangent drawn at A towards left = slope of the tangent drawn at A towards the right
The same tangent line towards left and right: meaning a unique tangent at the point
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:

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 definitely not differentiable at that point
If a function is not differentiable, then it may or may not be continuous at that point