How to Solve: Continuity and Differentiability (Chain Rule, Logarithmic/Implicit Differentiation) – IIT JEE Guide

How to Solve: Continuity and Differentiability (Chain Rule, Logarithmic/Implicit Differentiation) – IIT JEE Guide

Introduction Mastering continuity and differentiability—especially the Chain Rule, logarithmic, and implicit differentiation—can unlock 12-15 marks in IIT JEE (Main + Advanced). These concepts appear in every calculus problem, from limits to optimization, and are the backbone of 80% of JEE Advanced calculus questions. If you want to crack the top 100 ranks, you must solve these problems fast and error-free.

What You Need To Know First

Before diving in, ensure you’re 100% clear on these: 1. Limits & Continuity Basics – How to check if a function is continuous at a point (left-hand limit = right-hand limit = function value). 2. Basic Differentiation Rules – Power rule, product rule, quotient rule, and derivatives of standard functions (e.g., eˣ, ln x, sin x, cos x). 3. Composite Functions (f(g(x))) – How to identify inner and outer functions in nested expressions.

If any of these feel shaky, stop now and review them—this guide assumes you’re solid on these.

Key Vocabulary

Formulas To Know

1. Chain Rule (for Composite Functions)

Formula: If y = f(g(x)), then: dy/dx = f’(g(x)) × g’(x)

What it means: - f’(g(x)) = derivative of the outer function (evaluated at the inner function). - g’(x) = derivative of the inner function.

MEMORISE THIS – You’ll use it in every JEE calculus problem.

2. Implicit Differentiation

Formula: Differentiate both sides of the equation with respect to x, treating y as a function of x (so dy/dx appears).

Example: For x² + y² = 25, differentiate both sides: 2x + 2y (dy/dx) = 0 → Solve for dy/dx.

MEMORISE THIS – No formula, just a method.

3. Logarithmic Differentiation

Steps: 1. Take natural log (ln) of both sides. 2. Differentiate implicitly using d/dx [ln y] = (1/y) (dy/dx). 3. Solve for dy/dx.

When to use: - Functions like y = xˣ (variable base + variable exponent). - Complicated products/quotients (e.g., y = (x² + 1)ˣ).

MEMORISE THIS – It’s a lifesaver for tricky functions.

4. Differentiability Condition

Formula: A function f(x) is differentiable at x = a if: f’(a⁻) = f’(a⁺) (left-hand derivative = right-hand derivative).

MEMORISE THIS – Used to check differentiability at sharp points.

Step-by-Step Method

How to Solve Continuity & Differentiability Problems (IIT JEE Style)

Step 1: Check Continuity First

• For a given point x = a:
• Find f(a) (function value at a).
• Find limₓ→a⁻ f(x) (left-hand limit).
• Find limₓ→a⁺ f(x) (right-hand limit).

• If all three are equal, f(x) is continuous at x = a.

• If the function is piecewise:

• Check continuity at the joining point (usually x = 0 or x = 1).

Step 2: Check Differentiability

• If the function is continuous at x = a:
• Find f’(a⁻) (left-hand derivative).
• Find f’(a⁺) (right-hand derivative).

• If f’(a⁻) = f’(a⁺), the function is differentiable at x = a.

• If the function is not continuous at x = a:

• It cannot be differentiable there (differentiability ⇒ continuity).

Step 3: Differentiate Using the Right Rule

• For composite functions (e.g., sin(3x²)) → Chain Rule.
• For implicit functions (e.g., x² + y² = 1) → Implicit Differentiation.
• For y = xˣ or complicated products → Logarithmic Differentiation.

Step 4: Simplify & Solve for dy/dx

• After differentiating, isolate dy/dx (for implicit/logarithmic differentiation).
• Factor out dy/dx if needed.

Step 5: Check for Exam Traps

• Did you forget the Chain Rule on nested functions?
• Did you mishandle dy/dx in implicit differentiation?
• Did you assume differentiability without checking continuity first?

Worked Example (Using Exact Steps)

Problem: Let f(x) = |x - 1| + |x - 2|. Check: 1. Continuity at x = 1 and x = 2. 2. Differentiability at x = 1 and x = 2.

Step 1: Check Continuity at x = 1

1. Find f(1):
   f(1) = |1 - 1| + |1 - 2| = 0 + 1 = 1.

2. Left-hand limit (x → 1⁻):
   For x < 1, |x - 1| = 1 - x, |x - 2| = 2 - x.
   f(x) = (1 - x) + (2 - x) = 3 - 2x.
   limₓ→1⁻ f(x) = 3 - 2(1) = 1.

3. Right-hand limit (x → 1⁺):
   For 1 < x < 2, |x - 1| = x - 1, |x - 2| = 2 - x.
   f(x) = (x - 1) + (2 - x) = 1.
   limₓ→1⁺ f(x) = 1.

4. Conclusion:
   f(1) = limₓ→1⁻ f(x) = limₓ→1⁺ f(x) = 1 → Continuous at x = 1.

Step 2: Check Differentiability at x = 1

1. Left-hand derivative (x → 1⁻):
   f(x) = 3 - 2x → f’(x) = -2.
   f’(1⁻) = -2.

2. Right-hand derivative (x → 1⁺):
   f(x) = 1 → f’(x) = 0.
   f’(1⁺) = 0.

3. Conclusion:
   f’(1⁻) ≠ f’(1⁺) → Not differentiable at x = 1.

Step 3: Check Continuity at x = 2

1. Find f(2):
   f(2) = |2 - 1| + |2 - 2| = 1 + 0 = 1.

2. Left-hand limit (x → 2⁻):
   For 1 < x < 2, f(x) = 1 (from above).
   limₓ→2⁻ f(x) = 1.

3. Right-hand limit (x → 2⁺):
   For x > 2, |x - 1| = x - 1, |x - 2| = x - 2.
   f(x) = (x - 1) + (x - 2) = 2x - 3.
   limₓ→2⁺ f(x) = 2(2) - 3 = 1.

4. Conclusion:
   f(2) = limₓ→2⁻ f(x) = limₓ→2⁺ f(x) = 1 → Continuous at x = 2.

Step 4: Check Differentiability at x = 2

1. Left-hand derivative (x → 2⁻):
   f(x) = 1 → f’(x) = 0.
   f’(2⁻) = 0.

2. Right-hand derivative (x → 2⁺):
   f(x) = 2x - 3 → f’(x) = 2.
   f’(2⁺) = 2.

3. Conclusion:
   f’(2⁻) ≠ f’(2⁺) → Not differentiable at x = 2.

Final Answer: - f(x) is continuous at x = 1 and x = 2. - f(x) is not differentiable at x = 1 and x = 2.

What we did and why: - We checked continuity first because differentiability requires continuity. - We split the function into cases based on critical points (x = 1, 2). - We computed derivatives from both sides to check for smoothness (differentiability).

Worked Examples

Example 1 – Basic (Chain Rule)

Problem: Find dy/dx if y = sin(5x² + 3x).

Solution: 1. Identify inner and outer functions:
- Outer: sin(u), where u = 5x² + 3x.
- Inner: u = 5x² + 3x.

1. Differentiate outer function:
   d/dx [sin(u)] = cos(u) × du/dx.

2. Differentiate inner function:
   du/dx = 10x + 3.

3. Combine using Chain Rule:
   dy/dx = cos(5x² + 3x) × (10x + 3).

Final Answer: dy/dx = (10x + 3) cos(5x² + 3x).

What we did and why: - We applied the Chain Rule because sin(5x² + 3x) is a composite function. - We did not expand the final answer—JEE expects factored forms.

Example 2 – Medium (Implicit Differentiation)

Problem: Find dy/dx if x³ + y³ = 6xy.

Solution: 1. Differentiate both sides w.r.t. x:
d/dx [x³] + d/dx [y³] = d/dx [6xy].

1. Apply differentiation rules:
2. d/dx [x³] = 3x².
3. d/dx [y³] = 3y² (dy/dx) (Chain Rule).

4. d/dx [6xy] = 6y + 6x (dy/dx) (Product Rule).

5. Substitute back:
   3x² + 3y² (dy/dx) = 6y + 6x (dy/dx).

6. Collect dy/dx terms:
   3y² (dy/dx) - 6x (dy/dx) = 6y - 3x².

7. Factor out dy/dx:
   (dy/dx)(3y² - 6x) = 6y - 3x².

8. Solve for dy/dx:
   dy/dx = (6y - 3x²) / (3y² - 6x) = (2y - x²) / (y² - 2x).

Final Answer: dy/dx = (2y - x²) / (y² - 2x).

What we did and why: - We treated y as a function of x and used the Chain Rule. - We applied the Product Rule on 6xy. - We factored and simplified to match JEE answer formats.

Example 3 – Exam-Style (Logarithmic Differentiation)

Problem: Find dy/dx if y = (x² + 1)ˣ.

Solution: 1. Take natural log of both sides:
ln y = ln[(x² + 1)ˣ] = x ln(x² + 1).

1. Differentiate both sides w.r.t. x:
2. Left side: d/dx [ln y] = (1/y) (dy/dx).

3. Right side: d/dx [x ln(x² + 1)] = ln(x² + 1) + x × (2x)/(x² + 1) (Product Rule + Chain Rule).

4. Substitute back:
   (1/y) (dy/dx) = ln(x² + 1) + (2x²)/(x² + 1).

5. Multiply both sides by y:
   dy/dx = y [ln(x² + 1) + (2x²)/(x² + 1)].

6. Substitute y = (x² + 1)ˣ:
   dy/dx = (x² + 1)ˣ [ln(x² + 1) + (2x²)/(x² + 1)].

Final Answer: dy/dx = (x² + 1)ˣ [ln(x² + 1) + (2x²)/(x² + 1)].

What we did and why: - We used logarithmic differentiation because y has a variable base and exponent. - We applied the Product Rule on x ln(x² + 1). - We substituted back y at the end to match JEE answer formats.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 60-second survival guide for continuity and differentiability in JEE.

1. Continuity first, differentiability second. If a function isn’t continuous at a point, it cannot be differentiable there. Always check f(a), limₓ→a⁻, and limₓ→a⁺ first.

2. Chain Rule is your best friend. If you see sin(3x²), don’t just write cos(3x²)—multiply by the derivative of the inside (6x). Every nested function needs the Chain Rule.

3. Implicit differentiation? Differentiate both sides, treat y as a function of x, and always multiply by dy/dx when differentiating y terms.

4. Logarithmic differentiation for y = f(x)ᵍ⁽ˣ⁾. Take ln of both sides, differentiate, then solve for dy/dx. Never expand xˣ—it’s a trap!

5. Differentiability = smooth slope. If the left and right derivatives don’t match, the function isn’t differentiable at that point.

6. Watch for absolute values and piecewise functions. JEE loves testing |x| and functions defined in parts—check the joining points carefully.

You’ve got this. Now go crush that exam!