The Hook: From Speed to Acceleration
In Mission: Impossible - Dead Reckoning (2023), Ethan Hunt drives a car off a cliff. We know:
- Position $s(t)$ tells where the car is
- Velocity $v(t) = s'(t)$ tells how fast it’s moving
- Acceleration $a(t) = v'(t) = s''(t)$ tells how quickly the speed is changing
The second derivative $s''(t)$ is acceleration — the rate of change of rate of change!
Question: If $s(t) = 5t^2 + 20t$, what’s the acceleration?
$s'(t) = 10t + 20$ (velocity)
$s''(t) = 10$ (acceleration)
The car accelerates at a constant 10 m/s²!
This is the power of higher derivatives — they reveal deeper layers of change.
Interactive: Visualize Position, Velocity, Acceleration
See how the graphs of $s(t)$, $s'(t)$, and $s''(t)$ relate to each other!
What are Higher Derivatives?
The Derivative Chain
First Derivative: $f'(x) = \frac{df}{dx}$ — Rate of change of $f$
Second Derivative: $f''(x) = \frac{d}{dx}[f'(x)] = \frac{d^2f}{dx^2}$ — Rate of change of $f'$
Third Derivative: $f'''(x) = \frac{d}{dx}[f''(x)] = \frac{d^3f}{dx^3}$
nth Derivative: $f^{(n)}(x) = \frac{d^n f}{dx^n}$
Notation Systems
Multiple ways to denote higher derivatives:
| Order | Lagrange | Leibniz | Newton |
|---|---|---|---|
| 1st | $f'(x)$ | $\frac{dy}{dx}$ | $\dot{y}$ |
| 2nd | $f''(x)$ | $\frac{d^2y}{dx^2}$ | $\ddot{y}$ |
| 3rd | $f'''(x)$ | $\frac{d^3y}{dx^3}$ | $\dddot{y}$ |
| nth | $f^{(n)}(x)$ | $\frac{d^ny}{dx^n}$ | $y^{(n)}$ |
Note: $\frac{d^2y}{dx^2}$ does NOT mean $\left(\frac{dy}{dx}\right)^2$ — it’s the second derivative!
Computing Second Derivatives
Method: Differentiate Twice
Example 1: $f(x) = x^4$
$$f'(x) = 4x^3$$ $$f''(x) = \frac{d}{dx}(4x^3) = 12x^2$$Example 2: $y = \sin(2x)$
$$\frac{dy}{dx} = 2\cos(2x)$$ $$\frac{d^2y}{dx^2} = \frac{d}{dx}[2\cos(2x)] = -4\sin(2x)$$Example 3: $f(x) = e^{3x}$
$$f'(x) = 3e^{3x}$$ $$f''(x) = 9e^{3x}$$Higher Derivatives: General Patterns
Pattern 1: Polynomial Functions
For $f(x) = x^n$:
$$f'(x) = nx^{n-1}$$ $$f''(x) = n(n-1)x^{n-2}$$ $$f'''(x) = n(n-1)(n-2)x^{n-3}$$ $$f^{(n)}(x) = n!$$ $$f^{(n+1)}(x) = 0$$(and all higher derivatives are zero)
Example: $f(x) = x^5$
- $f'(x) = 5x^4$
- $f''(x) = 20x^3$
- $f'''(x) = 60x^2$
- $f^{(4)}(x) = 120x$
- $f^{(5)}(x) = 120$
- $f^{(6)}(x) = 0$
Pattern 2: Exponential Function
For $f(x) = e^{ax}$:
$$\boxed{f^{(n)}(x) = a^n e^{ax}}$$Example: $f(x) = e^{2x}$
- $f'(x) = 2e^{2x}$
- $f''(x) = 4e^{2x}$
- $f'''(x) = 8e^{2x}$
- $f^{(n)}(x) = 2^n e^{2x}$
Special case: $f(x) = e^x \Rightarrow f^{(n)}(x) = e^x$ (all derivatives equal to original!)
Pattern 3: Trigonometric Functions
For $f(x) = \sin(ax)$:
Pattern: $\sin \to \cos \to -\sin \to -\cos \to \sin$ (cycle of 4)
Example: $f(x) = \sin x$
- $f'(x) = \cos x = \sin(x + \frac{\pi}{2})$
- $f''(x) = -\sin x = \sin(x + \pi)$
- $f'''(x) = -\cos x = \sin(x + \frac{3\pi}{2})$
- $f^{(4)}(x) = \sin x = \sin(x + 2\pi)$
For $f(x) = \cos(ax)$:
Pattern: $\cos \to -\sin \to -\cos \to \sin \to \cos$ (cycle of 4)
Pattern 4: Logarithmic Function
For $f(x) = \ln x$:
$$f'(x) = \frac{1}{x} = x^{-1}$$ $$f''(x) = -x^{-2} = -\frac{1}{x^2}$$ $$f'''(x) = 2x^{-3} = \frac{2}{x^3}$$Second Derivatives in Different Forms
1. Implicit Functions
For $F(x, y) = 0$, we found $\frac{dy}{dx}$ by implicit differentiation.
To find $\frac{d^2y}{dx^2}$:
Method: Differentiate $\frac{dy}{dx}$ implicitly again!
Example: $x^2 + y^2 = 25$
We found: $\frac{dy}{dx} = -\frac{x}{y}$
Differentiate again:
$$\frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}$$ $$= -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}$$ $$= -\frac{y^2 + x^2}{y^3} = -\frac{25}{y^3}$$(since $x^2 + y^2 = 25$)
2. Parametric Functions
For $x = f(t)$, $y = g(t)$:
We know: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Second Derivative in Parametric Form:
$$\boxed{\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}}$$Steps:
- Find $\frac{dy}{dx}$ in terms of $t$
- Differentiate w.r.t. $t$
- Divide by $\frac{dx}{dt}$
Example: $x = t^2$, $y = t^3$
$$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$$ $$\frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$$ $$\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$$Leibniz’s Theorem for nth Derivative of Product
When finding nth derivative of a product $u(x) \cdot v(x)$:
Leibniz’s Theorem:
$$(uv)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(k)} v^{(n-k)}$$Expanded form:
$$(uv)^{(n)} = u^{(n)}v + \binom{n}{1}u^{(n-1)}v' + \binom{n}{2}u^{(n-2)}v'' + \ldots + uv^{(n)}$$Similar to binomial theorem: $(a + b)^n = \sum \binom{n}{k} a^k b^{n-k}$
Example
Find the 5th derivative of $y = x^2 e^x$
Let $u = x^2$, $v = e^x$
Derivatives of $u$:
- $u = x^2$
- $u' = 2x$
- $u'' = 2$
- $u''' = 0$ (and all higher)
Derivatives of $v$:
- $v = e^x$
- $v' = e^x$
- $v'' = e^x$
- (all derivatives are $e^x$)
Using Leibniz:
$$y^{(5)} = u^{(5)}v + 5u^{(4)}v' + 10u^{(3)}v'' + 10u''v''' + 5u'v^{(4)} + uv^{(5)}$$ $$= 0 + 0 + 0 + 10(2)e^x + 5(2x)e^x + x^2e^x$$ $$= 20e^x + 10xe^x + x^2e^x$$ $$= e^x(x^2 + 10x + 20)$$Physical Interpretations
In Kinematics (Physics)
Position: $s(t)$
Velocity: $v(t) = s'(t) = \frac{ds}{dt}$
Acceleration: $a(t) = v'(t) = s''(t) = \frac{d^2s}{dt^2}$
Jerk: $j(t) = a'(t) = s'''(t) = \frac{d^3s}{dt^3}$ (rate of change of acceleration)
Example: A particle moves with $s(t) = t^3 - 6t^2 + 9t$. Find velocity, acceleration, and jerk.
$$v(t) = s'(t) = 3t^2 - 12t + 9$$ $$a(t) = s''(t) = 6t - 12$$ $$j(t) = s'''(t) = 6$$At $t = 2$:
- Velocity: $v(2) = 12 - 24 + 9 = -3$ m/s (moving backward)
- Acceleration: $a(2) = 12 - 12 = 0$ m/s² (momentarily constant speed)
- Jerk: $j(2) = 6$ m/s³ (acceleration increasing)
In Curve Analysis
First Derivative: $f'(x)$ determines slope and monotonicity
- $f'(x) > 0$: Function increasing
- $f'(x) < 0$: Function decreasing
Second Derivative: $f''(x)$ determines concavity
- $f''(x) > 0$: Concave up (curve opens upward) ∪
- $f''(x) < 0$: Concave down (curve opens downward) ∩
Special Formulas and Results
nth Derivative Formulas
$(x^n)^{(n)} = n!$ (and $(x^n)^{(k)} = 0$ for $k > n$)
$(e^{ax})^{(n)} = a^n e^{ax}$
$(\sin ax)^{(n)} = a^n \sin(ax + \frac{n\pi}{2})$
$(\cos ax)^{(n)} = a^n \cos(ax + \frac{n\pi}{2})$
$(\ln x)^{(n)} = \frac{(-1)^{n-1}(n-1)!}{x^n}$
$(a^x)^{(n)} = a^x (\ln a)^n$
$\left(\frac{1}{ax+b}\right)^{(n)} = \frac{(-1)^n \cdot n! \cdot a^n}{(ax+b)^{n+1}}$
Memory Tricks
For $\sin x$ derivatives, remember the cycle:
$\sin \to \cos \to -\sin \to -\cos \to \sin$
Cos → Sin → Cos → Sin (with alternating signs)
Or think: “Sine Cycles: S→C→S→C, with + - - + pattern”
Leibniz’s theorem for $(uv)^{(n)}$ follows the same binomial coefficients as $(a+b)^n$!
Just replace powers with derivatives:
$$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ $$(uv)''' = u'''v + 3u''v' + 3u'v'' + uv'''$$“Differentiate, Divide by $\frac{dx}{d**T**}"$
$$\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$$Practice Problems
Level 1: Foundation (NCERT)
Question: Find $\frac{d^2y}{dx^2}$ if $y = x^4 + 3x^2 + 5$
Solution:
$$\frac{dy}{dx} = 4x^3 + 6x$$ $$\frac{d^2y}{dx^2} = 12x^2 + 6$$Question: If $y = \sin(3x)$, find $y''$
Solution:
$$y' = 3\cos(3x)$$ $$y'' = -9\sin(3x)$$Level 2: JEE Main
Question: Find the second derivative of $y = e^{2x}\cos x$
Solution (Product Rule):
$$y' = 2e^{2x}\cos x + e^{2x}(-\sin x) = e^{2x}(2\cos x - \sin x)$$ $$y'' = 2e^{2x}(2\cos x - \sin x) + e^{2x}(-2\sin x - \cos x)$$ $$= e^{2x}[4\cos x - 2\sin x - 2\sin x - \cos x]$$ $$= e^{2x}(3\cos x - 4\sin x)$$Question: If $y = x^3 \ln x$, find $y''$
Solution:
$$y' = 3x^2 \ln x + x^3 \cdot \frac{1}{x} = 3x^2\ln x + x^2$$ $$y'' = 6x\ln x + 3x^2 \cdot \frac{1}{x} + 2x = 6x\ln x + 3x + 2x$$ $$= 6x\ln x + 5x = x(6\ln x + 5)$$Question: If $x = at^2$, $y = 2at$, find $\frac{d^2y}{dx^2}$
Solution:
$$\frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 2a$$ $$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$$ $$\frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2}$$ $$\frac{d^2y}{dx^2} = \frac{-1/t^2}{2at} = -\frac{1}{2at^3}$$Level 3: JEE Advanced
Question: Find the 10th derivative of $y = \sin^2 x$
Solution:
Rewrite using double angle: $y = \frac{1 - \cos 2x}{2}$
$$y = \frac{1}{2} - \frac{1}{2}\cos 2x$$For $\cos 2x$, the 10th derivative follows the pattern:
$$(\cos 2x)^{(10)} = 2^{10} \cos(2x + 10 \cdot \frac{\pi}{2}) = 1024\cos(2x + 5\pi)$$ $$= 1024\cos(2x + \pi) = -1024\cos 2x$$Therefore:
$$y^{(10)} = 0 - \frac{1}{2}(-1024\cos 2x) = 512\cos 2x$$Question: Find the 5th derivative of $y = x^2 \sin x$ using Leibniz theorem
Solution:
Let $u = x^2$, $v = \sin x$
Derivatives of $u$: $u' = 2x$, $u'' = 2$, $u''' = 0$, …
Derivatives of $v$: $v = \sin x$, $v' = \cos x$, $v'' = -\sin x$, $v''' = -\cos x$, $v^{(4)} = \sin x$, $v^{(5)} = \cos x$
Using $(uv)^{(5)} = u^{(5)}v + 5u^{(4)}v' + 10u'''v'' + 10u''v''' + 5u'v^{(4)} + uv^{(5)}$:
$$y^{(5)} = 0 + 0 + 0 + 10(2)(-\cos x) + 5(2x)(\sin x) + x^2(\cos x)$$ $$= -20\cos x + 10x\sin x + x^2\cos x$$ $$= (x^2 - 20)\cos x + 10x\sin x$$Question: If $y = (x + \sqrt{1+x^2})^n$, prove that $(1+x^2)y'' + xy' = n^2y$
Solution:
Let $y = (x + \sqrt{1+x^2})^n$
$$y' = n(x + \sqrt{1+x^2})^{n-1} \left(1 + \frac{x}{\sqrt{1+x^2}}\right)$$ $$= n(x + \sqrt{1+x^2})^{n-1} \cdot \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}$$ $$= \frac{n(x + \sqrt{1+x^2})^n}{\sqrt{1+x^2}} = \frac{ny}{\sqrt{1+x^2}}$$So: $y'\sqrt{1+x^2} = ny$
Square both sides: $(y')^2(1+x^2) = n^2y^2$ … (*)
Differentiate (*):
$$2y'y''(1+x^2) + (y')^2(2x) = 2n^2yy'$$Divide by $2y'$:
$$y''(1+x^2) + xy' = n^2y$$$(1+x^2)y'' + xy' = n^2y$ ∎
Question: Find $\frac{d^{100}}{dx^{100}}(x^{99}e^x)$
Solution (Leibniz Theorem):
Let $u = x^{99}$, $v = e^x$
For $u = x^{99}$:
- $u^{(99)} = 99!$
- $u^{(100)} = 0$
For $v = e^x$: all derivatives are $e^x$
Using Leibniz:
$$(uv)^{(100)} = \sum_{k=0}^{100} \binom{100}{k} u^{(k)} v^{(100-k)}$$Since $u^{(k)} = 0$ for $k > 99$:
$$(uv)^{(100)} = \binom{100}{99} u^{(99)} v' + \binom{100}{100} u^{(100)} v$$ $$= 100 \cdot 99! \cdot e^x + 1 \cdot 0 \cdot e^x$$ $$= 100! \cdot e^x$$Common Mistakes to Avoid
Wrong: $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$ ✗
Correct: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ✓
$d^2y/dx^2$ means second derivative, NOT the square of first derivative!
Wrong: $\frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2}$ ✗
Correct: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$ ✓
You must differentiate $\frac{dy}{dx}$ (not $y$ directly), then divide by $\frac{dx}{dt}$!
When finding higher derivatives of products, don’t forget product rule at each step!
For $y = x^2e^x$:
- $y' = 2xe^x + x^2e^x$ ✓
- $y'' = (2xe^x)' + (x^2e^x)'$ ← Apply product rule to EACH term!
For $\sin x$ derivatives, the cycle is: $\sin \to \cos \to -\sin \to -\cos \to \sin$
Keep track of signs carefully! Use the formula $\sin^{(n)}(x) = \sin(x + n\pi/2)$ if unsure.
Quick Revision Box
| Function | 2nd Derivative | nth Derivative |
|---|---|---|
| $x^n$ | $n(n-1)x^{n-2}$ | $n!$ (for $n$th), $0$ (for higher) |
| $e^{ax}$ | $a^2e^{ax}$ | $a^n e^{ax}$ |
| $\sin(ax)$ | $-a^2\sin(ax)$ | $a^n\sin(ax + n\pi/2)$ |
| $\cos(ax)$ | $-a^2\cos(ax)$ | $a^n\cos(ax + n\pi/2)$ |
| $\ln x$ | $-\frac{1}{x^2}$ | $\frac{(-1)^{n-1}(n-1)!}{x^n}$ |
JEE Strategy Tips
Weightage: Higher derivatives appear in 1-2 JEE Main questions, 2-3 in JEE Advanced (often with physics applications).
Time-Saver: Memorize the standard nth derivative formulas — saves computation time!
Common Pattern: JEE loves asking nth derivative of products using Leibniz theorem, especially with $e^x$ and trig functions.
Trap Alert: Parametric second derivatives are tricky! Remember: differentiate $\frac{dy}{dx}$ w.r.t. $t$, then divide by $\frac{dx}{dt}$.
Physics Connection: Second derivative problems often have kinematics context (acceleration, jerk). Understand the physical meaning!
Exam Hack: For $y = P(x)e^{ax}$ where $P$ is polynomial degree $n$, the $(n+1)$th derivative will have no polynomial part, just exponential.
Teacher’s Summary
Higher derivatives reveal deeper patterns of change: velocity → acceleration → jerk in physics.
Standard formulas for $e^{ax}$, $\sin ax$, $\cos ax$, $\ln x$ save massive time — memorize them!
Leibniz theorem extends product rule to nth derivatives: $(uv)^{(n)} = \sum \binom{n}{k} u^{(k)}v^{(n-k)}$ (like binomial theorem).
Parametric second derivatives require differentiating $\frac{dy}{dx}$ w.r.t. $t$, then dividing by $\frac{dx}{dt}$ again.
Physical meaning: $f''(x) > 0$ means concave up, $f''(x) < 0$ means concave down — crucial for curve sketching!
“Derivatives are like layers of an onion — each layer reveals a new rate of change!”
Related Topics
Within Limits, Continuity & Differentiability
- Differentiation Rules — Foundation for higher derivatives
- Implicit Differentiation — Second derivatives of implicit functions
- Applications of Derivatives — Using second derivatives for concavity
- Curve Sketching — Second derivative test for inflection points
Applications
- Mean Value Theorems — Involving derivatives
- Maxima-Minima — Second derivative test
- Differential Equations — Equations with higher derivatives
Physics Connections
- Kinematics — Velocity, acceleration, jerk
- Simple Harmonic Motion — $a = -\omega^2 x$ (second derivative)
- Newton’s Laws — $F = ma = m\frac{d^2x}{dt^2}$