The Hook: Hidden Relationships
In 3 Idiots (2009), Rancho says “Don’t chase success, chase excellence and success will follow.”
Similarly, in the equation of a circle $x^2 + y^2 = 25$, we can’t directly solve for $y$ as a simple function of $x$ (we get $y = \pm\sqrt{25 - x^2}$).
Question: How do we find the slope of the tangent line to this circle at any point?
We need implicit differentiation — a technique that finds $\frac{dy}{dx}$ without explicitly solving for $y$!
This is the power of working with relationships hidden in equations.
Interactive: Visualize Implicit Curves
Plot implicit curves like $x^2 + y^2 = 25$ and see how the tangent slope changes!
What is Implicit Differentiation?
Explicit vs Implicit Functions
Explicit Function: $y$ is written explicitly in terms of $x$
- Example: $y = x^2 + 3x + 1$
- Easy to differentiate: $\frac{dy}{dx} = 2x + 3$
Implicit Function: $x$ and $y$ are mixed together in an equation
- Example: $x^2 + y^2 = 25$
- Can’t easily solve for $y$
The Technique
Key Idea: Differentiate both sides with respect to $x$, treating $y$ as a function of $x$.
Chain Rule for $y$ terms:
When differentiating $y^n$ with respect to $x$:
$$\frac{d}{dx}(y^n) = ny^{n-1} \cdot \frac{dy}{dx}$$Don’t forget the $\frac{dy}{dx}$ factor!
Step-by-Step Method
Example 1: Circle Equation
Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$
Step 1: Differentiate both sides w.r.t. $x$
$$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$$Step 2: Apply differentiation rules
$$2x + 2y\frac{dy}{dx} = 0$$(Remember: $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$ by chain rule!)
Step 3: Solve for $\frac{dy}{dx}$
$$2y\frac{dy}{dx} = -2x$$ $$\boxed{\frac{dy}{dx} = -\frac{x}{y}}$$Interpretation: At point $(3, 4)$ on the circle, slope = $-\frac{3}{4}$
Example 2: More Complex Equation
Find $\frac{dy}{dx}$ if $x^3 + y^3 = 6xy$
Differentiate both sides:
$$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$$(Used product rule on the right: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$)
Collect $\frac{dy}{dx}$ terms:
$$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$$ $$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$ $$\boxed{\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}}$$Common Patterns in Implicit Differentiation
Pattern 1: $y^n$ Terms
$$\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$$Examples:
- $\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$
- $\frac{d}{dx}(\sqrt{y}) = \frac{1}{2\sqrt{y}}\frac{dy}{dx}$
Pattern 2: Product of $x$ and $y$
$$\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$$ $$\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}$$Pattern 3: Trigonometric Functions of $y$
$$\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$$ $$\frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx}$$Pattern 4: Exponential/Logarithmic with $y$
$$\frac{d}{dx}(e^y) = e^y\frac{dy}{dx}$$ $$\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}$$Parametric Differentiation
What are Parametric Equations?
Instead of $y = f(x)$, both $x$ and $y$ are expressed in terms of a third variable (parameter) $t$:
$$x = f(t), \quad y = g(t)$$Example: Motion of a projectile
- $x = v_0t\cos\theta$ (horizontal position)
- $y = v_0t\sin\theta - \frac{1}{2}gt^2$ (vertical position)
Finding $\frac{dy}{dx}$ from Parametric Form
Parametric Derivative Formula:
$$\boxed{\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}}$$Provided $\frac{dx}{dt} \neq 0$
Intuition: By chain rule:
$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy/dt}{dx/dt}$$Example 1: Simple Parametric Curve
If $x = t^2$ and $y = t^3$, find $\frac{dy}{dx}$.
Solution:
$$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2$$ $$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$$To express in terms of $x$: Since $t = \sqrt{x}$,
$$\frac{dy}{dx} = \frac{3\sqrt{x}}{2}$$Example 2: Cycloid Curve
A cycloid is traced by a point on a rolling circle:
$$x = a(t - \sin t), \quad y = a(1 - \cos t)$$Find $\frac{dy}{dx}$.
Solution:
$$\frac{dx}{dt} = a(1 - \cos t)$$ $$\frac{dy}{dt} = a\sin t$$ $$\frac{dy}{dx} = \frac{a\sin t}{a(1 - \cos t)} = \frac{\sin t}{1 - \cos t}$$Simplify using half-angle formulas:
$$\sin t = 2\sin\frac{t}{2}\cos\frac{t}{2}, \quad 1 - \cos t = 2\sin^2\frac{t}{2}$$ $$\frac{dy}{dx} = \frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}} = \frac{\cos\frac{t}{2}}{\sin\frac{t}{2}} = \cot\frac{t}{2}$$Second Derivatives in Parametric Form
Second Derivative:
$$\boxed{\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}}$$Example
For $x = at^2$, $y = 2at$, find $\frac{d^2y}{dx^2}$.
First derivative:
$$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$$Second derivative:
$$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2}$$ $$\frac{d^2y}{dx^2} = \frac{-\frac{1}{t^2}}{2at} = -\frac{1}{2at^3}$$Special Cases and Applications
Case 1: Logarithmic Implicit Differentiation
For equations with products/quotients, take $\ln$ first!
Example: If $xy = e^{x-y}$, find $\frac{dy}{dx}$.
Method 1: Direct implicit differentiation (tedious)
Method 2: Take $\ln$ first (clever!)
$$\ln(xy) = \ln(e^{x-y})$$ $$\ln x + \ln y = x - y$$Differentiate:
$$\frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx}$$ $$\frac{dy}{dx}\left(\frac{1}{y} + 1\right) = 1 - \frac{1}{x}$$ $$\frac{dy}{dx} = \frac{1 - \frac{1}{x}}{\frac{1}{y} + 1} = \frac{y(x-1)}{x(y+1)}$$Case 2: Tangent and Normal to Curves
For implicit curve $F(x, y) = 0$ at point $(x_0, y_0)$:
Tangent Line:
$$y - y_0 = \frac{dy}{dx}\bigg|_{(x_0, y_0)} (x - x_0)$$Normal Line (perpendicular to tangent):
$$y - y_0 = -\frac{1}{\frac{dy}{dx}\bigg|_{(x_0, y_0)}} (x - x_0)$$Example: Find the tangent to $x^2 + y^2 = 25$ at $(3, 4)$.
We found earlier: $\frac{dy}{dx} = -\frac{x}{y}$
At $(3, 4)$: $\frac{dy}{dx} = -\frac{3}{4}$
Tangent: $y - 4 = -\frac{3}{4}(x - 3)$
Simplify: $3x + 4y = 25$
Case 3: Slope of Polar Curves
For polar curve $r = f(\theta)$:
$$x = r\cos\theta, \quad y = r\sin\theta$$Memory Tricks
When you see a $y$ term, think CHAIN rule!
$$\frac{d}{dx}(y^n) = ny^{n-1} \times \frac{dy}{dx}$$“Don’t forget dy/dx, it’s your friend!”
For parametric:
$$\frac{dy}{dx} = \frac{\text{derivative of } y}{\text{derivative of } x} = \frac{dy/dt}{dx/dt}$$“Divide the derivatives, both w.r.t. parameter!”
Practice Problems
Level 1: Foundation (NCERT)
Question: Find $\frac{dy}{dx}$ if $x^2 + y^2 = 16$
Solution:
Differentiate both sides:
$$2x + 2y\frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = -\frac{x}{y}$$Question: If $x = 3t$, $y = t^2$, find $\frac{dy}{dx}$
Solution:
$$\frac{dx}{dt} = 3, \quad \frac{dy}{dt} = 2t$$ $$\frac{dy}{dx} = \frac{2t}{3}$$Level 2: JEE Main
Question: Find $\frac{dy}{dx}$ if $xy + y^2 = \tan x$
Solution:
Differentiate:
$$y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = \sec^2 x$$ $$\frac{dy}{dx}(x + 2y) = \sec^2 x - y$$ $$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y}$$Question: If $x = a\cos^3\theta$, $y = a\sin^3\theta$, find $\frac{dy}{dx}$
Solution:
$$\frac{dx}{d\theta} = -3a\cos^2\theta\sin\theta$$ $$\frac{dy}{d\theta} = 3a\sin^2\theta\cos\theta$$ $$\frac{dy}{dx} = \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\frac{\sin\theta}{\cos\theta} = -\tan\theta$$Question: Find the equation of tangent to $x^2 + y^2 = 13$ at point $(2, 3)$
Solution:
Implicit differentiation: $\frac{dy}{dx} = -\frac{x}{y}$
At $(2, 3)$: $\frac{dy}{dx} = -\frac{2}{3}$
Tangent equation:
$$y - 3 = -\frac{2}{3}(x - 2)$$ $$3y - 9 = -2x + 4$$ $$2x + 3y = 13$$Level 3: JEE Advanced
Question: If $x\sqrt{1+y} + y\sqrt{1+x} = 0$, prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$
Solution:
Rewrite: $x\sqrt{1+y} = -y\sqrt{1+x}$
Square both sides:
$$x^2(1+y) = y^2(1+x)$$ $$x^2 + x^2y = y^2 + xy^2$$ $$x^2 - y^2 = xy^2 - x^2y = xy(y - x)$$ $$(x-y)(x+y) = -xy(x-y)$$If $x \neq y$, divide by $(x-y)$:
$$x + y = -xy$$Differentiate:
$$1 + \frac{dy}{dx} = -y - x\frac{dy}{dx}$$ $$\frac{dy}{dx}(1 + x) = -1 - y$$From $x + y = -xy$, we get $y = -\frac{x}{1+x}$
So $1 + y = \frac{1}{1+x}$
$$\frac{dy}{dx} = \frac{-1-y}{1+x} = \frac{-(1+y)}{1+x} = -\frac{1}{(1+x)^2}$$Hence proved. ∎
Question: If $x = a(\theta + \sin\theta)$, $y = a(1 - \cos\theta)$, find $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{2}$
Solution:
$$\frac{dx}{d\theta} = a(1 + \cos\theta), \quad \frac{dy}{d\theta} = a\sin\theta$$ $$\frac{dy}{dx} = \frac{a\sin\theta}{a(1+\cos\theta)} = \frac{\sin\theta}{1+\cos\theta}$$For second derivative:
$$\frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{(1+\cos\theta)\cos\theta - \sin\theta(-\sin\theta)}{(1+\cos\theta)^2}$$ $$= \frac{\cos\theta + \cos^2\theta + \sin^2\theta}{(1+\cos\theta)^2} = \frac{\cos\theta + 1}{(1+\cos\theta)^2} = \frac{1}{1+\cos\theta}$$ $$\frac{d^2y}{dx^2} = \frac{\frac{1}{1+\cos\theta}}{a(1+\cos\theta)} = \frac{1}{a(1+\cos\theta)^2}$$At $\theta = \frac{\pi}{2}$: $\cos\frac{\pi}{2} = 0$
$$\frac{d^2y}{dx^2} = \frac{1}{a(1+0)^2} = \frac{1}{a}$$Question: If $y = y(x)$ satisfies $x\cos y + y\cos x = \pi$, find $y''(0)$.
Solution:
At $x = 0$: $0 + y\cos 0 = \pi \Rightarrow y(0) = \pi$
Differentiate: $\cos y - x\sin y \cdot y' + y' \cos x - y\sin x = 0$
At $x = 0$, $y = \pi$:
$$\cos\pi - 0 + y'(0)\cos 0 - \pi \sin 0 = 0$$ $$-1 + y'(0) = 0 \Rightarrow y'(0) = 1$$Differentiate again:
$$-\sin y \cdot y' - [\sin y \cdot y' + x\sin y \cdot (y')^2 + x\cos y \cdot y''] + [y'' \cos x - y' \sin x] - [y'\sin x + y\cos x] = 0$$At $x = 0$, $y = \pi$, $y' = 1$:
$$0 - 0 + y''(0) - 1 - 0 = 0$$ $$y''(0) = 1$$Common Mistakes to Avoid
Wrong: $\frac{d}{dx}(y^2) = 2y$ ✗
Correct: $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ ✓
Always apply chain rule to $y$ terms!
Wrong: $\frac{d}{dx}(xy) = y$ ✗
Correct: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ ✓
Use product rule!
Wrong: $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$ ✗
Correct: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ✓
"$dy$ on top, $dx$ on bottom!"
Quick Revision Box
| Concept | Formula | Key Point |
|---|---|---|
| Implicit Differentiation | Differentiate both sides w.r.t. $x$ | Don’t forget $\frac{dy}{dx}$ on $y$ terms |
| $y^n$ terms | $\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$ | Chain rule always! |
| $xy$ terms | $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ | Product rule |
| Parametric | $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | Divide derivatives |
| Second derivative | $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$ | Differentiate w.r.t. $t$, then divide |
JEE Strategy Tips
Weightage: Implicit and parametric differentiation appear in 2-3 questions every JEE.
Time-Saver: For tangent/normal problems, implicit differentiation is often faster than solving for $y$ explicitly!
Common Pattern: JEE loves mixing implicit + parametric. Be ready to convert between forms.
Trap Alert: Always check if $\frac{dx}{dt} = 0$ at any point in parametric — derivative is undefined there (vertical tangent)!
Advanced Tip: For complex implicit equations, taking $\ln$ before differentiating can simplify massively.
Exam Hack: In parametric second derivatives, express $\frac{dy}{dx}$ purely in terms of $t$ before differentiating again.
Teacher’s Summary
Implicit differentiation finds $\frac{dy}{dx}$ without solving for $y$ — essential for circles, ellipses, and complex curves.
Golden rule: Always apply chain rule to $y$ terms: $\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$
Parametric curves express both $x$ and $y$ in terms of $t$. Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Second derivatives in parametric form require differentiating $\frac{dy}{dx}$ w.r.t. $t$, then dividing by $\frac{dx}{dt}$ again.
Applications include tangent/normal to curves, physics (motion), and polar coordinates.
“Implicit differentiation: when $y$ is shy and won’t come out explicitly!”
Related Topics
Within Limits, Continuity & Differentiability
- Differentiation Rules — Foundation for implicit/parametric techniques
- Higher Derivatives — Extending to second and higher orders
- Applications of Derivatives — Using derivatives for tangents, normals
Mathematical Foundations
- Functions — Understanding implicit relations
- Trigonometry — Trig identities in parametric forms
- Coordinate Geometry — Curves and tangents
Applications
- Curve Sketching — Using derivatives to analyze curves
- Differential Equations — Implicit differentiation in DEs
Physics Connections
- Kinematics — Parametric equations for projectile motion
- Circular Motion — Polar coordinate derivatives