The Hook: Speedometer Science
Shah Rukh Khan is chasing villains in a high-speed car. At exactly 10:00:00 AM, what is his speed?
To calculate speed, you need: Speed = Distance/Time
But at a single instant, time interval = 0, so you get $\frac{\text{distance}}{0}$ — undefined!
Yet the speedometer shows 120 km/h. How?
This is the miracle of derivatives — the mathematical tool that finds instantaneous rates of change when direct calculation fails. The derivative is the “speedometer” of mathematics!
Interactive: Explore Tangent Lines
Plot $f(x) = x^2$ and observe how the tangent line’s slope changes as you move along the curve. That slope is the derivative!
The Core Concept
What is a Derivative?
The derivative of $f(x)$ at $x = a$ is:
$$\boxed{f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}}$$Alternate form:
$$\boxed{f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}}$$In simple terms: The derivative is the slope of the tangent line to the graph at $x = a$.
Geometric Meaning
$$\frac{f(a + h) - f(a)}{h}$$is the slope of the secant line joining $(a, f(a))$ and $(a+h, f(a+h))$.
As $h \to 0$, the secant line becomes the tangent line, and the slope becomes $f'(a)$.
If $s(t)$ is position at time $t$:
- $\frac{s(t + h) - s(t)}{h}$ = Average velocity
- $s'(t) = \lim_{h \to 0} \frac{s(t + h) - s(t)}{h}$ = Instantaneous velocity
Derivative = Instantaneous rate of change!
What is Differentiability?
Definition
A function $f(x)$ is differentiable at $x = a$ if:
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$exists (and is finite).
In simple terms: The function has a well-defined tangent line at $x = a$.
Left Hand Derivative (LHD) & Right Hand Derivative (RHD)
Left Hand Derivative:
$$\boxed{f'(a^-) = \lim_{h \to 0^-} \frac{f(a + h) - f(a)}{h}}$$(Approaching from the left)
Right Hand Derivative:
$$\boxed{f'(a^+) = \lim_{h \to 0^+} \frac{f(a + h) - f(a)}{h}}$$(Approaching from the right)
The Golden Rule
$$\boxed{f \text{ is differentiable at } a \iff f'(a^-) = f'(a^+) \text{ (both finite)}}$$If LHD ≠ RHD, the function is not differentiable at $a$.
Differentiability Implies Continuity
The Theorem
$$\boxed{\text{If } f \text{ is differentiable at } a \text{, then } f \text{ is continuous at } a}$$Proof Idea:
If $f'(a)$ exists, then:
$$\lim_{h \to 0} [f(a + h) - f(a)] = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \cdot h = f'(a) \cdot 0 = 0$$So $\lim_{x \to a} f(x) = f(a)$ → Continuous!
But the Converse is FALSE!
$$\boxed{\text{Continuous does NOT imply differentiable}}$$Classic Example: $f(x) = |x|$ at $x = 0$
- Continuous at $x = 0$ ✓
- Sharp corner at $x = 0$ → Not differentiable ✗
Differentiable → Continuous ✓
Continuous → Differentiable ✗
JEE loves to test this distinction!
Relationship Diagram
Differentiable ⟹ Continuous ⟹ Limit exists
But:
Limit exists ⟹̸ Continuous
Continuous ⟹̸ Differentiable
Example Flow:
- $f(x) = |x|$ is continuous everywhere but not differentiable at $x = 0$
- $f(x) = \frac{x^2 - 1}{x - 1}$ has a limit at $x = 1$ but is not continuous there
- $f(x) = \sin x$ is differentiable everywhere → continuous everywhere → limit exists everywhere
When is a Function NOT Differentiable?
Case 1: Sharp Corner (Cusp)
Example: $f(x) = |x|$ at $x = 0$
$$f(x) = \begin{cases} -x & x < 0 \\ x & x \geq 0 \end{cases}$$LHD: $f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$
RHD: $f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$
LHD ≠ RHD → Not differentiable at $x = 0$.
Case 2: Vertical Tangent
Example: $f(x) = x^{1/3}$ at $x = 0$
$$f'(0) = \lim_{h \to 0} \frac{h^{1/3} - 0}{h} = \lim_{h \to 0} \frac{1}{h^{2/3}} = \infty$$Derivative is infinite → Not differentiable.
(The tangent line is vertical, slope = ∞)
Case 3: Discontinuity
If $f$ is discontinuous at $x = a$, it cannot be differentiable there.
Example: $f(x) = \lfloor x \rfloor$ (greatest integer function)
Discontinuous at all integers → Not differentiable at integers.
Case 4: Sharp Turn in Piecewise Functions
Example:
$$f(x) = \begin{cases} x^2 & x < 1 \\ 2x - 1 & x \geq 1 \end{cases}$$Check at $x = 1$:
LHD: $f'(1^-) = \lim_{h \to 0^-} \frac{(1+h)^2 - 1}{h} = \lim_{h \to 0} \frac{2h + h^2}{h} = 2$
RHD: $f'(1^+) = \lim_{h \to 0^+} \frac{2(1+h) - 1 - 1}{h} = \lim_{h \to 0} \frac{2h}{h} = 2$
LHD = RHD = 2 → Differentiable at $x = 1$!
(Smooth join, no sharp corner)
Notation for Derivatives
Multiple ways to denote the derivative of $y = f(x)$:
| Notation | Meaning |
|---|---|
| $f'(x)$ | Lagrange notation |
| $\frac{dy}{dx}$ | Leibniz notation |
| $\frac{df}{dx}$ | Leibniz (alternate) |
| $Df(x)$ | Operator notation |
| $y'$ | Simplified |
All mean the same thing!
Basic Derivatives from First Principles
Derivative of $x^n$
$$f(x) = x^n$$ $$f'(x) = \lim_{h \to 0} \frac{(x + h)^n - x^n}{h}$$Using binomial theorem or the standard limit $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$:
$$\boxed{f'(x) = nx^{n-1}}$$Examples:
- $f(x) = x^3 \Rightarrow f'(x) = 3x^2$
- $f(x) = x \Rightarrow f'(x) = 1$
- $f(x) = 1 \Rightarrow f'(x) = 0$
Derivative of $\sin x$
$$f(x) = \sin x$$ $$f'(x) = \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h}$$Using $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$:
$$= \lim_{h \to 0} \frac{2\cos(x + h/2)\sin(h/2)}{h}$$ $$= \lim_{h \to 0} \cos(x + h/2) \cdot \frac{\sin(h/2)}{h/2}$$ $$= \cos x \cdot 1 = \cos x$$ $$\boxed{\frac{d}{dx}(\sin x) = \cos x}$$Derivative of $e^x$
$$f(x) = e^x$$ $$f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} e^x \cdot \frac{e^h - 1}{h}$$Using $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$:
$$\boxed{\frac{d}{dx}(e^x) = e^x}$$(The only function equal to its own derivative!)
Table of Basic Derivatives
| Function $f(x)$ | Derivative $f'(x)$ |
|---|---|
| $c$ (constant) | $0$ |
| $x$ | $1$ |
| $x^n$ | $nx^{n-1}$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\csc x$ | $-\csc x \cot x$ |
| $e^x$ | $e^x$ |
| $a^x$ | $a^x \ln a$ |
| $\ln x$ | $\frac{1}{x}$ |
| $\log_a x$ | $\frac{1}{x \ln a}$ |
| $\sqrt{x}$ | $\frac{1}{2\sqrt{x}}$ |
Inverse Trigonometric Derivatives
| Function | Derivative |
|---|---|
| $\sin^{-1} x$ | $\frac{1}{\sqrt{1 - x^2}}$ |
| $\cos^{-1} x$ | $\frac{-1}{\sqrt{1 - x^2}}$ |
| $\tan^{-1} x$ | $\frac{1}{1 + x^2}$ |
| $\cot^{-1} x$ | $\frac{-1}{1 + x^2}$ |
| $\sec^{-1} x$ | $\frac{1}{ |
| $\csc^{-1} x$ | $\frac{-1}{ |
Memory Tricks & Patterns
The “Co-Function” Pattern
For trig derivatives, co-functions get a minus sign:
$$\frac{d}{dx}(\cos x) = -\sin x$$ $$\frac{d}{dx}(\cot x) = -\csc^2 x$$ $$\frac{d}{dx}(\csc x) = -\csc x \cot x$$Mnemonic: “Co-functions are Cold (negative)”
The “Square Pattern”
$$\frac{d}{dx}(\tan x) = \sec^2 x$$ $$\frac{d}{dx}(\cot x) = -\csc^2 x$$Mnemonic: “Tan and Cot derivatives are squared”
The “Product Pattern”
$$\frac{d}{dx}(\sec x) = \sec x \tan x$$ $$\frac{d}{dx}(\csc x) = -\csc x \cot x$$Mnemonic: “Sec and Csc derivatives are products (function × neighbor)”
Practice Problems
Level 1: Foundation (NCERT)
Question: Find the derivative of $f(x) = x^5$ from first principles.
Solution:
$$f'(x) = \lim_{h \to 0} \frac{(x + h)^5 - x^5}{h}$$Using $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$:
$$f'(x) = 5x^4$$Question: Is $f(x) = |x|$ differentiable at $x = 0$?
Solution:
$$f(x) = \begin{cases} -x & x < 0 \\ x & x \geq 0 \end{cases}$$LHD: $f'(0^-) = -1$
RHD: $f'(0^+) = 1$
LHD ≠ RHD → Not differentiable at $x = 0$.
Level 2: JEE Main
Question:
$$f(x) = \begin{cases} x^2 \sin(1/x) & x \neq 0 \\ 0 & x = 0 \end{cases}$$Is $f$ differentiable at $x = 0$?
Solution:
$$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h}$$ $$= \lim_{h \to 0} h \sin(1/h)$$Since $-1 \leq \sin(1/h) \leq 1$:
$$-|h| \leq h\sin(1/h) \leq |h|$$By squeeze theorem: $\lim_{h \to 0} h\sin(1/h) = 0$
Differentiable at $x = 0$, with $f'(0) = 0$.
Question:
$$f(x) = \begin{cases} x^2 & x \leq 1 \\ ax + b & x > 1 \end{cases}$$Find $a$ and $b$ for $f$ to be differentiable at $x = 1$.
Solution: For continuity at $x = 1$:
$$\lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x)$$ $$1 = a + b \quad \ldots (1)$$For differentiability at $x = 1$:
$$f'(1^-) = f'(1^+)$$LHD: $f'(1^-) = 2(1) = 2$
RHD: $f'(1^+) = a$
So $a = 2$.
From (1): $b = 1 - 2 = -1$
Answer: $a = 2, b = -1$
Level 3: JEE Advanced
Question: Show that $f(x) = x|x|$ is differentiable at $x = 0$ and find $f'(0)$.
Solution:
$$f(x) = \begin{cases} -x^2 & x < 0 \\ x^2 & x \geq 0 \end{cases}$$LHD:
$$f'(0^-) = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h) = 0$$RHD:
$$f'(0^+) = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0$$LHD = RHD = 0 → Differentiable at $x = 0$, with $f'(0) = 0$.
Question: If $f(x) = x^n$, show that $f'(x) = nx^{n-1}$ using first principles.
Solution:
$$f'(x) = \lim_{h \to 0} \frac{(x + h)^n - x^n}{h}$$Let $x + h = t$. As $h \to 0$, $t \to x$.
$$= \lim_{t \to x} \frac{t^n - x^n}{t - x}$$Using the standard limit $\lim_{t \to a} \frac{t^n - a^n}{t - a} = na^{n-1}$:
$$f'(x) = nx^{n-1}$$Question:
$$f(x) = \begin{cases} x^2 + 2x & x < 0 \\ 2x & 0 \leq x < 1 \\ ax^2 + bx + c & x \geq 1 \end{cases}$$Find $a$, $b$, $c$ for $f$ to be differentiable everywhere.
Solution: At $x = 0$:
Continuity: $\lim_{x \to 0^-} (x^2 + 2x) = 0 = f(0) = 0$ ✓
LHD: $f'(0^-) = 2(0) + 2 = 2$
RHD: $f'(0^+) = 2$
Differentiable at $x = 0$ ✓
At $x = 1$:
Continuity: $2(1) = a + b + c$
$$a + b + c = 2 \quad \ldots (1)$$LHD: $f'(1^-) = 2$
RHD: $f'(1^+) = 2a + b$
Differentiability: $2a + b = 2 \quad \ldots (2)$
Also need continuity of derivative (for smoothness):
$$f''(1^-) = f''(1^+) \Rightarrow 0 = 2a$$ $$a = 0$$From (2): $b = 2$
From (1): $c = 0$
Answer: $a = 0, b = 2, c = 0$
Common Mistakes to Avoid
Wrong: $f(x) = |x|$ is continuous everywhere, so it’s differentiable everywhere. ✗
Correct: $f(x) = |x|$ is continuous everywhere but not differentiable at $x = 0$ (sharp corner).
LHL/RHL are for limits (continuity).
LHD/RHD are for derivatives (differentiability).
They use similar symbols but test different things!
For differentiability at a point in a piecewise function:
- Check continuity first (LHL = RHL = f(a))
- Then check differentiability (LHD = RHD)
If continuity fails, differentiability automatically fails!
Quick Revision Box
| Concept | Formula/Rule |
|---|---|
| Derivative definition | $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ |
| Differentiability test | LHD = RHD (both finite) |
| Key implication | Differentiable ⟹ Continuous |
| NOT differentiable when | Sharp corner, vertical tangent, discontinuity |
| Power rule | $(x^n)' = nx^{n-1}$ |
| Sin rule | $(\sin x)' = \cos x$ |
| Exponential rule | $(e^x)' = e^x$ |
JEE Strategy Tips
Weightage: Differentiability appears in 2-3 questions in JEE Main, 3-4 in JEE Advanced (often combined with continuity).
Time-Saver: For piecewise functions, if continuity fails, don’t waste time checking differentiability — it’s automatically not differentiable!
Common Pattern: JEE loves giving piecewise functions with parameters. Template:
- Set LHL = RHL = f(a) for continuity
- Calculate LHD and RHD separately
- Set LHD = RHD for differentiability
Trap Alert: $|x|$ at $x = 0$ is the most common counter-example. Memorize it!
Advanced Tip: If asked to prove differentiability from first principles, use the alternate form $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ — sometimes easier!
Teacher’s Summary
Derivative = Instantaneous rate of change = Slope of tangent line. It’s the “speedometer” of math.
Differentiable ⟹ Continuous, but continuous does NOT imply differentiable. $|x|$ is the classic counter-example.
LHD = RHD (finite) is the test for differentiability. Sharp corners, vertical tangents, and discontinuities all cause failure.
For piecewise functions: Check continuity first (saves time), then check LHD = RHD at boundary points.
Master the basic derivatives — power rule, trig, exponential, log. They’re the building blocks for all differentiation.
“Differentiability is continuity with smoothness!”
Related Topics
Within Limits, Continuity & Differentiability
- Continuity — Differentiability implies continuity
- Differentiation Rules — Product, quotient, chain rules
- Standard Limits — Used in derivative definitions
Mathematical Foundations
- Functions Basics — Understanding function behavior
- Trigonometry — Trig function derivatives
Applications
- Applications of Derivatives — Maxima, minima, rate of change
- Integral Calculus — Derivatives and integrals are inverses
- Differential Equations — Equations involving derivatives
Physics Connections
- Kinematics — Velocity = derivative of position
- Newton’s Laws — Force involves derivatives of momentum
- SHM — Acceleration = second derivative