How to Solve: Application of Derivatives (IIT JEE Main + Advanced)

How to Solve: Application of Derivatives (IIT JEE Main + Advanced)

Introduction

"Mastering derivatives isn’t just about passing JEE—it’s about cracking 15-20 marks in your exam, from tangent lines to optimization problems that appear in every paper. One wrong step, and you lose 4-6 marks. Get this right, and you’re already ahead of 80% of students."

What You Need To Know First

1. Basic differentiation rules (power, product, quotient, chain rule).
2. Equation of a line (slope-intercept form, point-slope form).
3. Critical points and second derivative test (basics of maxima/minima).

Key Vocabulary

Formulas To Know

1. Equation of Tangent & Normal

• Tangent at ( x = a ): [ y - f(a) = f'(a)(x - a) ]
• ( f(a) ) = y-coordinate at ( x = a )

• ( f'(a) ) = slope of tangent at ( x = a ) MEMORISE THIS

• Normal at ( x = a ): [ y - f(a) = -\frac{1}{f'(a)}(x - a) ] (Slope of normal = negative reciprocal of tangent slope) MEMORISE THIS

2. Maxima & Minima

• First Derivative Test:
• If ( f'(x) ) changes from + to -, ( x ) is a local max.

• If ( f'(x) ) changes from - to +, ( x ) is a local min. MEMORISE THIS

• Second Derivative Test:

• If ( f''(a) > 0 ), ( x = a ) is a local min.
• If ( f''(a) < 0 ), ( x = a ) is a local max.
• If ( f''(a) = 0 ), test fails (use first derivative test). MEMORISE THIS

3. Rate of Change

• Related Rates: [ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} ] (Chain rule for rates) MEMORISE THIS

4. Mean Value Theorem (MVT)

• If ( f(x) ) is continuous on ([a, b]) and differentiable on ((a, b)), then: [ f'(c) = \frac{f(b) - f(a)}{b - a} ] for some ( c \in (a, b) ). GIVEN ON EXAM SHEET (but understand it!)

Step-by-Step Method

1. Tangent & Normal Lines

Steps: 1. Find ( f(a) ): Plug ( x = a ) into ( f(x) ) to get the point ((a, f(a))). 2. Find ( f'(a) ): Differentiate ( f(x) ) and plug in ( x = a ) to get the slope. 3. Write tangent equation: Use point-slope form ( y - f(a) = f'(a)(x - a) ). 4. Write normal equation: Use slope ( m = -\frac{1}{f'(a)} ) and same point.

Example: Find the tangent and normal to ( f(x) = x^3 ) at ( x = 2 ).

Solution: 1. ( f(2) = 8 ) → Point: ((2, 8)) 2. ( f'(x) = 3x^2 ) → ( f'(2) = 12 ) 3. Tangent: ( y - 8 = 12(x - 2) ) → ( y = 12x - 16 ) 4. Normal: ( y - 8 = -\frac{1}{12}(x - 2) ) → ( y = -\frac{1}{12}x + \frac{49}{6} )

2. Maxima & Minima

Steps: 1. Find ( f'(x) ): Differentiate the function. 2. Find critical points: Solve ( f'(x) = 0 ) or where ( f'(x) ) is undefined. 3. Test critical points:
- First Derivative Test: Check sign changes of ( f'(x) ) around critical points.
- Second Derivative Test: Plug critical points into ( f''(x) ). 4. Check endpoints (if domain is closed). 5. Compare values to find absolute max/min.

Example: Find local maxima/minima of ( f(x) = x^3 - 3x^2 ).

Solution: 1. ( f'(x) = 3x^2 - 6x ) 2. ( 3x^2 - 6x = 0 ) → ( x = 0, 2 ) 3. First Derivative Test:
- For ( x = 0 ): ( f'(x) ) changes from + to - → Local max.
- For ( x = 2 ): ( f'(x) ) changes from - to + → Local min. 4. Second Derivative Test:
- ( f''(x) = 6x - 6 )
- ( f''(0) = -6 < 0 ) → Local max at ( x = 0 ).
- ( f''(2) = 6 > 0 ) → Local min at ( x = 2 ).

3. Rate of Change (Related Rates)

Steps: 1. Identify variables: What is changing? (e.g., radius, volume, distance) 2. Write an equation relating the variables. 3. Differentiate implicitly with respect to time ( t ). 4. Plug in known values and solve for the unknown rate.

Example: A balloon’s radius increases at ( 2 ) cm/s. How fast is its volume increasing when ( r = 5 ) cm?

Solution: 1. Variables: ( V ) (volume), ( r ) (radius), ( t ) (time). 2. Equation: ( V = \frac{4}{3}\pi r^3 ) 3. Differentiate: ( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} ) 4. Plug in: ( \frac{dr}{dt} = 2 ), ( r = 5 )
( \frac{dV}{dt} = 4\pi (25)(2) = 200\pi ) cm³/s

4. Mean Value Theorem (MVT)

Steps: 1. Check conditions: Is ( f(x) ) continuous on ([a, b]) and differentiable on ((a, b))? 2. Compute ( \frac{f(b) - f(a)}{b - a} ). 3. Set ( f'(c) = ) that value and solve for ( c ).

Example: Verify MVT for ( f(x) = x^2 ) on ([1, 3]).

Solution: 1. ( f(x) ) is continuous and differentiable everywhere. 2. ( \frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4 ) 3. ( f'(x) = 2x )
( 2c = 4 ) → ( c = 2 ) (which is in ((1, 3))).

Worked Examples

Example 1 – Basic (Tangent Line)

Problem: Find the tangent to ( y = \ln x ) at ( x = 1 ).

Solution: 1. ( f(1) = \ln 1 = 0 ) → Point: ((1, 0)) 2. ( f'(x) = \frac{1}{x} ) → ( f'(1) = 1 ) 3. Tangent: ( y - 0 = 1(x - 1) ) → ( y = x - 1 )

What we did and why: We used the point-slope form of a line, where the slope is the derivative at the given point.

Example 2 – Medium (Maxima/Minima)

Problem: Find the absolute maximum of ( f(x) = x^3 - 6x^2 + 9x ) on ([0, 4]).

Solution: 1. ( f'(x) = 3x^2 - 12x + 9 ) 2. ( 3x^2 - 12x + 9 = 0 ) → ( x = 1, 3 ) 3. First Derivative Test:
- ( x = 1 ): ( f'(x) ) changes from + to - → Local max.
- ( x = 3 ): ( f'(x) ) changes from - to + → Local min. 4. Check endpoints:
- ( f(0) = 0 )
- ( f(4) = 16 ) 5. Compare values:
- ( f(1) = 4 )
- ( f(3) = 0 )
- Absolute max = 16 at ( x = 4 ).

What we did and why: We checked critical points and endpoints because the domain was closed. The absolute max was at an endpoint, not a critical point!

Example 3 – Exam-Style (Related Rates)

Problem: A ladder 10 m long leans against a wall. If the bottom slides away at 1 m/s, how fast is the top sliding down when the bottom is 6 m from the wall?

Solution: 1. Variables: ( x ) (distance from wall), ( y ) (height on wall), ( t ) (time). 2. Equation: ( x^2 + y^2 = 100 ) 3. Differentiate: ( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 ) 4. Plug in: ( \frac{dx}{dt} = 1 ), ( x = 6 ) → ( y = 8 )
( 2(6)(1) + 2(8) \frac{dy}{dt} = 0 )
( \frac{dy}{dt} = -\frac{6}{8} = -0.75 ) m/s

What we did and why: We used the Pythagorean theorem and implicit differentiation. The negative sign means the top is sliding down.

Common Mistakes

Exam Traps

1-Minute Recap

"Listen up—this is your last-minute cheat sheet for derivatives in JEE:

1. Tangent/Normal: Find ( f(a) ), ( f'(a) ), then plug into point-slope form. Normal slope is the negative reciprocal.
2. Maxima/Minima: Find critical points, test with first or second derivative, always check endpoints if the domain is closed.
3. Rate of Change: Write an equation, differentiate implicitly, plug in known rates.
4. Mean Value Theorem: Check continuity and differentiability, then solve ( f'(c) = \frac{f(b) - f(a)}{b - a} ).

Common traps? Forgetting endpoints, mixing up tangent/normal slopes, and misapplying MVT. Double-check your work, and you’ll nail these 15-20 marks. Good luck!