The Hook: The Multiplication Mystery
In 12th Fail (2023), Manoj studies hard every day. His knowledge grows, and his confidence grows too.
Question: How fast is his total success (knowledge × confidence) growing?
If knowledge is function $u(t)$ and confidence is $v(t)$, success = $u(t) \times v(t)$.
Can we just multiply the derivatives? Is $(uv)' = u'v'$?
Try it: $u = x$, $v = x$, so $uv = x^2$.
- LHS: $(x^2)' = 2x$
- RHS: $(x)' \cdot (x)' = 1 \cdot 1 = 1$
They’re not equal! So how do we differentiate products?
This is where the Product Rule saves the day!
Interactive: Derivative as Slope Visualizer
This interactive tool lets you see what a derivative really means. Watch how the secant line (through two points) becomes the tangent line as the distance $h$ approaches zero. This is the definition of derivative in action!
How to use:
- Select a function from the dropdown
- Move the “Point a” slider to explore different locations on the curve
- Adjust the “h” slider to see how the secant slope approaches the tangent slope
- Click “Watch h -> 0” to see the limit animation
- Toggle visibility of secant, tangent, and derivative curves
Key insight: When the secant slope (orange dashed) matches the tangent slope (green solid), you’ve found the derivative!
Interactive: Explore Function Transformations
Try plotting composite functions like $\sin(x^2)$ and see how they differ from $\sin x \cdot x^2$!
The Core Concept
Why We Need Rules
Computing derivatives from first principles every time is tedious.
Differentiation rules let us find derivatives of complex functions instantly by breaking them down into simpler parts.
The Big Three Rules:
- Product Rule — For multiplying functions
- Quotient Rule — For dividing functions
- Chain Rule — For composing functions
Rule 1: The Product Rule
Statement
If $u(x)$ and $v(x)$ are differentiable, then:
$$\boxed{\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)}$$Short form: $(uv)' = u'v + uv'$
In words: “Derivative of first × second + first × derivative of second”
Proof Sketch
$$\frac{d}{dx}(uv) = \lim_{h \to 0} \frac{u(x+h)v(x+h) - u(x)v(x)}{h}$$Add and subtract $u(x+h)v(x)$:
$$= \lim_{h \to 0} \frac{u(x+h)v(x+h) - u(x+h)v(x) + u(x+h)v(x) - u(x)v(x)}{h}$$ $$= \lim_{h \to 0} \left[u(x+h) \cdot \frac{v(x+h) - v(x)}{h} + v(x) \cdot \frac{u(x+h) - u(x)}{h}\right]$$ $$= u(x) \cdot v'(x) + v(x) \cdot u'(x)$$Examples
Example 1: $f(x) = x^2 \sin x$
Let $u = x^2$, $v = \sin x$.
$$f'(x) = (x^2)' \sin x + x^2 (\sin x)'$$ $$= 2x \sin x + x^2 \cos x$$Example 2: $f(x) = e^x (x^3 + 2x)$
$$f'(x) = (e^x)' (x^3 + 2x) + e^x (x^3 + 2x)'$$ $$= e^x (x^3 + 2x) + e^x (3x^2 + 2)$$ $$= e^x (x^3 + 3x^2 + 2x + 2)$$$(uv)' = $ FIRST’ × SECOND + FIRST × SECOND'
Or think: “FS’ + SF’” (First Second-prime plus Second First-prime)
Extension: Product of Three Functions
For $f(x) = u(x) \cdot v(x) \cdot w(x)$:
$$\boxed{(uvw)' = u'vw + uv'w + uvw'}$$Pattern: Differentiate one at a time, keeping others constant, then add!
Rule 2: The Quotient Rule
Statement
If $u(x)$ and $v(x)$ are differentiable and $v(x) \neq 0$, then:
$$\boxed{\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{[v(x)]^2}}$$Short form: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
In words: “Bottom times derivative of top minus top times derivative of bottom, all over bottom squared”
Proof Sketch
Use product rule on $u = \frac{u}{v} \cdot v$:
$$u' = \left(\frac{u}{v}\right)' \cdot v + \frac{u}{v} \cdot v'$$Solving for $\left(\frac{u}{v}\right)'$:
$$\left(\frac{u}{v}\right)' = \frac{u' - \frac{u}{v} \cdot v'}{v} = \frac{u'v - uv'}{v^2}$$Examples
Example 1: $f(x) = \frac{x^2}{x + 1}$
Let $u = x^2$, $v = x + 1$.
$$f'(x) = \frac{(x^2)'(x+1) - x^2(x+1)'}{(x+1)^2}$$ $$= \frac{2x(x+1) - x^2(1)}{(x+1)^2}$$ $$= \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$$Example 2: $f(x) = \frac{\sin x}{x}$
$$f'(x) = \frac{(\sin x)' \cdot x - \sin x \cdot (x)'}{x^2}$$ $$= \frac{x\cos x - \sin x}{x^2}$$Think of a ladder leaning against a wall:
LOw × dHIgh - HIgh × dLOw, all over LO-LO
(Bottom × derivative of top - top × derivative of bottom, over bottom squared)
Or the cleaner version: “Bottom-Top’ minus Top-Bottom’, over Bottom-Bottom”
Rule 3: The Chain Rule
Statement
If $y = f(u)$ and $u = g(x)$, then:
$$\boxed{\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}}$$Alternate form: If $y = f(g(x))$:
$$\boxed{\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)}$$In words: “Derivative of outer function (at inner) × derivative of inner function”
Intuition
Think of a chain of events:
- $x$ changes → $u = g(x)$ changes → $y = f(u)$ changes
Rate of change of $y$ w.r.t. $x$ = (rate of $y$ w.r.t. $u$) × (rate of $u$ w.r.t. $x$)
Just like: km/hour = (km/miles) × (miles/hour) — the “miles” cancel!
Examples
Example 1: $y = \sin(x^2)$
Let $u = x^2$, so $y = \sin u$.
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos u \cdot 2x = 2x\cos(x^2)$$Example 2: $y = e^{3x}$
Let $u = 3x$, so $y = e^u$.
$$\frac{dy}{dx} = e^u \cdot 3 = 3e^{3x}$$Example 3: $y = \sqrt{x^2 + 1}$
Let $u = x^2 + 1$, so $y = u^{1/2}$.
$$\frac{dy}{dx} = \frac{1}{2}u^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$$Think of Ethan Hunt in Mission: Impossible defusing a bomb. To stop the explosion (change in $y$), he must:
- Cut the right wire (change $u$)
- Which requires the right tool (change $x$)
The chain of actions: tool → wire → explosion. Chain rule connects them!
Nested Chain Rule
For $y = f(g(h(x)))$:
$$\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$$Example: $y = \sin(\sqrt{x^2 + 1})$
$$\frac{dy}{dx} = \cos(\sqrt{x^2 + 1}) \cdot \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{x\cos(\sqrt{x^2 + 1})}{\sqrt{x^2 + 1}}$$Combining the Rules
When to Use Which?
| Function Type | Rule to Use |
|---|---|
| $u(x) \cdot v(x)$ | Product rule |
| $\frac{u(x)}{v(x)}$ | Quotient rule |
| $f(g(x))$ | Chain rule |
| $u(x) \cdot v(x) / w(x)$ | Product + Quotient |
| $f(g(h(x)))$ | Nested chain rule |
Complex Example
$$f(x) = \frac{x^2 \sin(x^3)}{e^x}$$Step 1: Quotient rule (outer structure)
Let $u = x^2 \sin(x^3)$, $v = e^x$.
$$f'(x) = \frac{u'e^x - u \cdot e^x}{(e^x)^2} = \frac{u' - u}{e^x}$$Step 2: Product rule for $u$
$u = x^2 \cdot \sin(x^3)$
$$u' = 2x\sin(x^3) + x^2 \cdot (\sin(x^3))'$$Step 3: Chain rule for $(\sin(x^3))'$
$$(\sin(x^3))' = \cos(x^3) \cdot 3x^2$$Combine:
$$u' = 2x\sin(x^3) + x^2 \cdot 3x^2\cos(x^3) = 2x\sin(x^3) + 3x^4\cos(x^3)$$ $$f'(x) = \frac{2x\sin(x^3) + 3x^4\cos(x^3) - x^2\sin(x^3)}{e^x}$$Advanced Techniques
1. Logarithmic Differentiation
When to use: Products/quotients of many terms, or variable exponents.
Method: Take $\ln$ of both sides, then differentiate.
Example: $y = x^x$ (variable base and exponent!)
Take $\ln$: $\ln y = \ln(x^x) = x \ln x$
Differentiate both sides:
$$\frac{1}{y} \cdot \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$$ $$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$$Example 2: $y = \frac{x^3 \sqrt{x+1}}{(x-2)^2}$
Take $\ln$:
$$\ln y = 3\ln x + \frac{1}{2}\ln(x+1) - 2\ln(x-2)$$Differentiate:
$$\frac{1}{y} \frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)} - \frac{2}{x-2}$$ $$\frac{dy}{dx} = y\left[\frac{3}{x} + \frac{1}{2(x+1)} - \frac{2}{x-2}\right]$$2. Implicit Differentiation
When to use: Equation is not in form $y = f(x)$ (e.g., $x^2 + y^2 = 1$).
Method: Differentiate both sides w.r.t. $x$, treating $y$ as a function of $x$.
Example: $x^2 + y^2 = 25$
Differentiate:
$$2x + 2y\frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = -\frac{x}{y}$$Example 2: $xy + \sin y = 1$
Differentiate:
$$y + x\frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = 0$$ $$\frac{dy}{dx}(x + \cos y) = -y$$ $$\frac{dy}{dx} = \frac{-y}{x + \cos y}$$3. Parametric Differentiation
When to use: $x$ and $y$ both expressed in terms of parameter $t$.
Formula:
$$\boxed{\frac{dy}{dx} = \frac{dy/dt}{dx/dt}}$$Example: $x = t^2$, $y = t^3$
$$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2$$ $$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$$Memory Tricks & Patterns
Product Rule Mnemonic
“First times Second-prime Plus Second times First-prime”
Or think: “FS’ + SF’”
Quotient Rule Mnemonic
“Low-dHigh minus High-dLow, over Low-Low”
Or: “Bottom-Top’ minus Top-Bottom’, over Bottom²”
Or sing to the tune: “LO-dHI, HI-dLO, square the bottom and away we go!”
Chain Rule Mnemonic
“Derivative of Outside × Derivative of Inside”
Think: “Outside-Inside-Prime” or “OIP”
Pattern Recognition
| Pattern | Quick Formula |
|---|---|
| $(ax + b)^n$ | $an(ax + b)^{n-1}$ |
| $e^{ax}$ | $ae^{ax}$ |
| $\sin(ax)$ | $a\cos(ax)$ |
| $\ln(ax + b)$ | $\frac{a}{ax + b}$ |
Practice Problems
Level 1: Foundation (NCERT)
Question: Differentiate $f(x) = x^3 \sin x$
Solution (Product Rule):
$$f'(x) = (x^3)' \sin x + x^3 (\sin x)'$$ $$= 3x^2 \sin x + x^3 \cos x$$Question: Find $\frac{d}{dx}\left(\frac{x^2}{x+1}\right)$
Solution (Quotient Rule):
$$\frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{2x(x+1) - x^2 \cdot 1}{(x+1)^2}$$ $$= \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$$Question: Differentiate $y = \sin(3x)$
Solution (Chain Rule):
$$\frac{dy}{dx} = \cos(3x) \cdot 3 = 3\cos(3x)$$Level 2: JEE Main
Question: Find $\frac{dy}{dx}$ if $y = e^{x^2}$
Solution (Chain Rule):
$$\frac{dy}{dx} = e^{x^2} \cdot 2x = 2xe^{x^2}$$Question: Differentiate $f(x) = \frac{\sin x}{x^2}$
Solution (Quotient Rule):
$$f'(x) = \frac{(\sin x)' \cdot x^2 - \sin x \cdot (x^2)'}{(x^2)^2}$$ $$= \frac{x^2 \cos x - 2x\sin x}{x^4} = \frac{x\cos x - 2\sin x}{x^3}$$Question: Find $\frac{dy}{dx}$ if $x^2 + y^2 = 100$
Solution (Implicit Differentiation):
$$2x + 2y\frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = -\frac{x}{y}$$Level 3: JEE Advanced
Question: Differentiate $y = x^{\sin x}$
Solution (Logarithmic Differentiation):
$$\ln y = \sin x \cdot \ln x$$Differentiate both sides:
$$\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}$$ $$\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x}\right)$$Question: If $x = a(t - \sin t)$ and $y = a(1 - \cos t)$, find $\frac{dy}{dx}$.
Solution (Parametric Differentiation):
$$\frac{dx}{dt} = a(1 - \cos t), \quad \frac{dy}{dt} = a\sin t$$ $$\frac{dy}{dx} = \frac{a\sin t}{a(1 - \cos t)} = \frac{\sin t}{1 - \cos t}$$Simplify using $\sin t = 2\sin(t/2)\cos(t/2)$ and $1 - \cos t = 2\sin^2(t/2)$:
$$\frac{dy}{dx} = \frac{2\sin(t/2)\cos(t/2)}{2\sin^2(t/2)} = \frac{\cos(t/2)}{\sin(t/2)} = \cot(t/2)$$Question: If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}}$ to infinity, find $\frac{dy}{dx}$.
Solution: Since the pattern repeats infinitely, we have:
$$y = \sqrt{\sin x + y}$$Square both sides:
$$y^2 = \sin x + y$$Differentiate implicitly:
$$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}$$ $$\frac{dy}{dx}(2y - 1) = \cos x$$ $$\frac{dy}{dx} = \frac{\cos x}{2y - 1}$$Question: Find $\frac{d^2y}{dx^2}$ if $y = e^{ax}\sin(bx)$
Solution: First derivative (Product Rule):
$$\frac{dy}{dx} = ae^{ax}\sin(bx) + be^{ax}\cos(bx) = e^{ax}[a\sin(bx) + b\cos(bx)]$$Second derivative (Product Rule again):
$$\frac{d^2y}{dx^2} = ae^{ax}[a\sin(bx) + b\cos(bx)] + e^{ax}[ab\cos(bx) - b^2\sin(bx)]$$ $$= e^{ax}[a^2\sin(bx) + ab\cos(bx) + ab\cos(bx) - b^2\sin(bx)]$$ $$= e^{ax}[(a^2 - b^2)\sin(bx) + 2ab\cos(bx)]$$Common Mistakes to Avoid
Wrong: $(uv)' = u'v'$ ✗
Correct: $(uv)' = u'v + uv'$ ✓
The derivatives do NOT multiply directly!
Wrong: $\left(\frac{u}{v}\right)' = \frac{uv' - u'v}{v^2}$ ✗
Correct: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ ✓
Top-prime comes first!
Wrong: $\frac{d}{dx}[\sin(x^2)] = \cos(x^2)$ ✗
Correct: $\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$ ✓
Don’t forget to multiply by the derivative of the inside!
When differentiating $y^2$ w.r.t. $x$:
Wrong: $\frac{d}{dx}(y^2) = 2y$ ✗
Correct: $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ ✓
Always include $\frac{dy}{dx}$ when differentiating $y$ terms!
Quick Revision Box
| Rule | Formula | When to Use |
|---|---|---|
| Product | $(uv)' = u'v + uv'$ | Functions multiplied |
| Quotient | $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ | Functions divided |
| Chain | $[f(g(x))]' = f'(g(x)) \cdot g'(x)$ | Composite functions |
| Logarithmic | Take $\ln$, then differentiate | Complex products, $y^x$, $x^y$, $x^x$ |
| Implicit | Differentiate both sides w.r.t. $x$ | Equation not solved for $y$ |
| Parametric | $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | Both $x$, $y$ functions of $t$ |
JEE Strategy Tips
Weightage: Differentiation rules appear in every calculus problem — they’re fundamental!
Time-Saver: Memorize the mnemonics! They save precious seconds:
- Product: “FS’ + SF'”
- Quotient: “LO-dHI minus HI-dLO over LO-LO”
- Chain: “Outside-Inside-Prime”
Common Pattern: JEE Main loves chain rule with standard functions. Recognize patterns like $(ax + b)^n$ → derivative = $an(ax + b)^{n-1}$
Advanced Trick: For complex fractions, sometimes logarithmic differentiation is faster than quotient rule!
Implicit Differentiation: Always appears in JEE Advanced. Practice until automatic!
Exam Hack: If you see $x^x$ or $x^{\sin x}$, immediately think logarithmic differentiation.
Teacher’s Summary
The Big Three — Product, Quotient, Chain — are your differentiation toolkit. Master them, and you can differentiate anything.
Product Rule: $u'v + uv'$ (derivative of first × second + first × derivative of second)
Quotient Rule: $\frac{u'v - uv'}{v^2}$ (top-prime bottom minus top bottom-prime, over bottom squared)
Chain Rule: Derivative of outside × derivative of inside. For nested functions, work from outside to inside.
Advanced techniques — Logarithmic (for variable exponents), Implicit (for unsolved $y$), Parametric (for $t$-based) — extend your power to any function.
“Differentiation rules turn calculus from a marathon into a sprint!”
Related Topics
Within Limits, Continuity & Differentiability
- Differentiability - Foundation for these rules
- Higher Derivatives - Second and higher order derivatives
- Implicit Differentiation - For implicit functions
- Applications of Derivatives - Using derivatives for maxima, minima
- Standard Limits - Used in derivative proofs
- Maxima and Minima - Optimization using derivatives
Mathematical Foundations
- Trigonometric Identities - Trig function derivatives
- Inverse Trigonometry - Inverse trig derivatives
Applications
- Indefinite Integrals - Derivatives and integrals are inverses
- Differential Equations - Solving equations with derivatives
- Tangent and Normal to Conics - Finding tangent lines
Physics Connections
- Kinematics - Chain rule for velocity/acceleration
- Simple Harmonic Motion - Differentiating $\sin$, $\cos$
- Current Electricity - Rate of change of current