Differentiability

1. Concept Overview

A function is said to be differentiable at a point if its derivative exists at that point.

This means the rate of change of the function is well-defined and unique at that point.

2. Mathematical Definition

A function [f(x)] is differentiable at [x=a] if the following limit exists and is finite:

[f'(a)][=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}]

This is the first principle (definition of derivative).

Differentiability from Left & Right

We check:

Left-hand derivative (LHD):

[\lim_{h\to 0^-}\dfrac{f(a+h)-f(a)}{h}]

Right-hand derivative (RHD):

[\lim_{h\to 0^+}\dfrac{f(a+h)-f(a)}{h}]

If LHD = RHD = finite, the function is differentiable at (x=a).
If not equal or infinite → not differentiable.

3. Important Result to Remember

If a function is differentiable at a point, then it must be continuous at that point.
But the converse is NOT always true (example: [f(x)=|x|] is continuous at 0 but not differentiable at 0).

We will explore this deeply with examples.

4. Graphical Meaning

A function is differentiable at a point if the graph has a unique tangent with:

✔ no sharp corners
✔ no cusps
✔ no breaks
✔ no vertical tangent

Example of a non-differentiable sharp corner: [f(x)=|x|] at [(x=0)]

5. Quick Example (First Principle)

Find derivative of [f(x)=x^2] using definition:

[f'(x)=\lim_{h\to 0}\dfrac{(x+h)^2-x^2}{h}]
[= \lim_{h\to 0}\dfrac{x^2+2xh+h^2-x^2}{h}]
[= \lim_{h\to 0}\dfrac{2xh+h^2}{h}]
[= \lim_{h\to 0}(2x+h)]
[= 2x]

6. Key Takeaways

7. Conceptual Questions with Solutions

8. FAQ / Common Misconceptions

9. Practice Questions (Step-by-Step Solutions — 10 minimum)

Question 1.

Check the differentiability of [f(x)=|x-2|] at [(x=2)].

Step-by-Step Solution:

1. Break into piecewise form:
   If [x ≥ 2], [f(x)=x-2]
   If [x < 2], [f(x)=2-x]
2. Left-hand derivative at [x=2]:
   [ \dfrac{d}{dx}(2-x) = -1 ]
3. Right-hand derivative at [x=2]:
   [ \dfrac{d}{dx}(x-2) = +1 ]
4. LHD ≠ RHD
   [ -1 ≠ +1 ]

Conclusion:
Not differentiable at [x=2] (corner point).

Question 2.

Check if [f(x)=x^2] is differentiable at [(x=0)].

Step-by-Step Solution:

1. [f(x)=x^2] is a polynomial function.
2. Polynomial functions are differentiable for all real x.
3. Derivative: [f'(x)=2x]
4. At [x=0], [f'(0)=0]

Conclusion:
Differentiable at all x, including [x=0].

Question 3.

Check differentiability of
[f(x)=\begin{cases}
x^2, & x ≤ 1 \
2x-1, & x > 1
\end{cases}] at [(x=1)].

Step-by-Step Solution:

1. Check continuity:
   Left limit at [1] → [1^2 = 1]
   Right limit at [1] → [2(1)-1 = 1]
   [f(1)=1] ✔ Continuous
2. Left derivative:
   [\dfrac{d}{dx}(x^2)=2x → 2(1)=2]
3. Right derivative:
   [\dfrac{d}{dx}(2x-1)=2]
   Right derivative at [1] → [2]
4. LHD = RHD = 2

Conclusion:
Differentiable at [x=1].

Question 4.

Find whether [f(x)=\dfrac{1}{x}] is differentiable at [(x=0)].

Step-by-Step Solution:

1. [f(x)] is not defined at [x=0].
2. No function value ⇒ No continuity.
3. If not continuous → Cannot be differentiable.

Conclusion:
Not differentiable at [x=0] (infinite discontinuity).

Question 5.

Check differentiability of [f(x)=x^{1/3}] at [(x=0)].

Step-by-Step Solution:

1. Derivative:
   [f'(x)=\dfrac{1}{3}x^{-2/3}]
2. At [x=0]:
   Derivative becomes infinite (division by zero).
3. Infinite derivative ⇒ derivative does not exist.

Conclusion:
Not differentiable at [x=0] (vertical tangent).

Summary of Concepts Used