Differentiability of Implicit Functions and Functions Defined by Parametric Equations
Master advanced differentiation techniques for JEE Main & Advanced with step-by-step methods and solved examples.
Why These Topics Are Crucial for JEE
Implicit functions and parametric equations appear frequently in JEE problems because they test your conceptual understanding of differentiation beyond basic formulas. Mastering these techniques gives you:
- Ability to differentiate complex relationships not explicitly solved for y
- Tools to handle real-world applications in physics and engineering
- Foundation for understanding related rates and optimization
- Edge in solving advanced calculus problems efficiently
🎯 Quick Navigation
1. Understanding Implicit Functions
What Are Implicit Functions?
An implicit function is defined by an equation where the dependent variable is not isolated on one side. Instead, the relationship between variables is given implicitly.
Explicit Function
$y = f(x)$
Example: $y = x^2 + 3x - 5$
y is explicitly defined in terms of x
Implicit Function
$F(x, y) = 0$
Example: $x^2 + y^2 - 25 = 0$
Relationship between x and y is implicit
Common Implicit Functions in JEE
Circle: $x^2 + y^2 = r^2$
Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Trigonometric: $\sin(x+y) = xy$
Exponential: $e^{x+y} = x^2y$
Logarithmic: $\ln(x^2 + y^2) = 2x$
2. Implicit Differentiation Method
The Step-by-Step Process
Step 1: Differentiate Both Sides
Differentiate both sides of the equation with respect to x, treating y as a function of x.
Step 2: Apply Chain Rule
When differentiating terms containing y, multiply by $\frac{dy}{dx}$ (chain rule).
Step 3: Solve for dy/dx
Collect all terms containing $\frac{dy}{dx}$ on one side and solve for it.
Example 1: Basic Implicit Differentiation
Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$
Solution:
Step 1: Differentiate both sides with respect to x:
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$
Step 2: Apply chain rule to y²:
$2x + 2y\frac{dy}{dx} = 0$
Step 3: Solve for $\frac{dy}{dx}$:
$2y\frac{dy}{dx} = -2x$
$\frac{dy}{dx} = -\frac{x}{y}$
Example 2: Advanced Implicit Differentiation
Find $\frac{dy}{dx}$ for $\sin(x+y) + \cos(x-y) = 1$
Solution:
Step 1: Differentiate both sides:
$\frac{d}{dx}[\sin(x+y)] + \frac{d}{dx}[\cos(x-y)] = 0$
Step 2: Apply chain rule:
$\cos(x+y)\cdot(1+\frac{dy}{dx}) - \sin(x-y)\cdot(1-\frac{dy}{dx}) = 0$
Step 3: Expand and collect terms:
$\cos(x+y) + \cos(x+y)\frac{dy}{dx} - \sin(x-y) + \sin(x-y)\frac{dy}{dx} = 0$
Step 4: Solve for $\frac{dy}{dx}$:
$\frac{dy}{dx}[\cos(x+y) + \sin(x-y)] = \sin(x-y) - \cos(x+y)$
$\frac{dy}{dx} = \frac{\sin(x-y) - \cos(x+y)}{\cos(x+y) + \sin(x-y)}$
💡 Key Insight for Implicit Differentiation
Remember that $\frac{d}{dx}(y^n) = n y^{n-1} \frac{dy}{dx}$ and $\frac{d}{dx}(f(y)) = f'(y) \frac{dy}{dx}$
3. Understanding Parametric Functions
What Are Parametric Functions?
A parametric function defines a relationship between variables using a third parameter (usually t or θ). Both x and y are expressed in terms of this parameter.
Parametric Form
$x = f(t)$, $y = g(t)$
where t is the parameter
Common Parametric Functions in JEE
Circle: $x = r\cos\theta$, $y = r\sin\theta$
Ellipse: $x = a\cos\theta$, $y = b\sin\theta$
Parabola: $x = at^2$, $y = 2at$
Hyperbola: $x = a\sec\theta$, $y = b\tan\theta$
Cycloid: $x = a(\theta - \sin\theta)$, $y = a(1 - \cos\theta)$
Astroid: $x = a\cos^3\theta$, $y = a\sin^3\theta$
⚠️ Important Note
Parametric representation is often more convenient for describing curves, especially when dealing with motion or geometric properties.
4. Parametric Differentiation Method
The Chain Rule Approach
Fundamental Formula
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$
provided $\frac{dx}{dt} \neq 0$
Step 1: Find dx/dt and dy/dt
Differentiate both x and y with respect to the parameter t.
Step 2: Apply the Formula
Use the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Step 3: Simplify
Express the result in simplest form, either in terms of t or x and y.
Example 3: Basic Parametric Differentiation
Find $\frac{dy}{dx}$ for $x = 2\cos\theta$, $y = 3\sin\theta$
Solution:
Step 1: Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:
$\frac{dx}{d\theta} = -2\sin\theta$, $\frac{dy}{d\theta} = 3\cos\theta$
Step 2: Apply the formula:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3\cos\theta}{-2\sin\theta} = -\frac{3}{2}\cot\theta$
Step 3: Express in terms of x and y (optional):
Since $\cos\theta = \frac{x}{2}$ and $\sin\theta = \frac{y}{3}$:
$\frac{dy}{dx} = -\frac{3}{2} \cdot \frac{x/2}{y/3} = -\frac{9x}{4y}$
Example 4: Second Derivative Parametric
Find $\frac{d^2y}{dx^2}$ for $x = a(\theta - \sin\theta)$, $y = a(1 - \cos\theta)$
Solution:
Step 1: Find first derivatives:
$\frac{dx}{d\theta} = a(1 - \cos\theta)$, $\frac{dy}{d\theta} = a\sin\theta$
$\frac{dy}{dx} = \frac{a\sin\theta}{a(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta}$
Step 2: Simplify using trigonometric identities:
$\frac{dy}{dx} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = \cot(\theta/2)$
Step 3: Find second derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\cot(\theta/2)\right) \cdot \frac{d\theta}{dx}$
$= -\frac{1}{2}\csc^2(\theta/2) \cdot \frac{1}{a(1 - \cos\theta)}$
Step 4: Simplify:
$\frac{d^2y}{dx^2} = -\frac{1}{2a(1 - \cos\theta)\sin^2(\theta/2)}$
Since $1 - \cos\theta = 2\sin^2(\theta/2)$:
$\frac{d^2y}{dx^2} = -\frac{1}{4a\sin^4(\theta/2)}$
💡 Formula for Second Derivative
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$
This formula is essential for finding higher order derivatives of parametric functions.
⚠️ Common Mistakes to Avoid
In Implicit Differentiation:
- Forgetting to multiply by $\frac{dy}{dx}$ when differentiating y terms
- Not applying chain rule correctly for composite functions
- Mishandling product rule and quotient rule with y terms
- Algebraic errors when solving for $\frac{dy}{dx}$
In Parametric Differentiation:
- Using $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$ (reversed)
- Forgetting that $\frac{dx}{dt} \neq 0$ is required
- Errors in finding higher order derivatives
- Not simplifying trigonometric expressions
📝 Practice Problems
Test your understanding with these JEE-level problems:
1. Find $\frac{dy}{dx}$ for $x^3 + y^3 = 3axy$
2. Find $\frac{d^2y}{dx^2}$ for $x = a\cos^3\theta$, $y = a\sin^3\theta$
3. If $e^{x+y} = xy$, find $\frac{dy}{dx}$ at point (1,1)
🚀 Quick Revision Guide
Implicit Differentiation
- Differentiate both sides w.r.t. x
- Use chain rule: $\frac{d}{dx}f(y) = f'(y)\frac{dy}{dx}$
- Solve for $\frac{dy}{dx}$ algebraically
- Check your answer by substituting known points
Parametric Differentiation
- Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
- Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
- For second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) / \frac{dx}{dt}$
- Simplify using trigonometric identities when possible
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