Higher-Order Derivatives: Mastering the nth Derivative
From basic patterns to Leibniz rule - complete guide to finding nth derivatives for JEE success.
Why Higher-Order Derivatives Matter in JEE
Higher-order derivatives appear in every JEE paper - from simple pattern recognition to complex Leibniz rule applications. Mastering this topic gives you:
- Quick 2-3 marks in MCQs through pattern recognition
- Foundation for Taylor/Maclaurin series and expansion problems
- Edge in integer/numerical answer type questions
- Better understanding of curve behavior and optimization
🎯 Quick Navigation
1. Fundamental Patterns & Formulas
Polynomial Functions
General Formula:
If $f(x) = x^m$, then:
Using factorial notation:
Example 1:
Find the 4th derivative of $f(x) = x^5$
Solution: $f^{(4)}(x) = \frac{5!}{(5-4)!}x^{5-4} = \frac{120}{1}x^1 = 120x$
Example 2:
Find the 7th derivative of $f(x) = x^3$
Solution: Since $n = 7 > 3 = m$, $f^{(7)}(x) = 0$
Exponential Functions
Key Formulas:
• $f(x) = e^{ax} \Rightarrow f^{(n)}(x) = a^n e^{ax}$
• $f(x) = a^x \Rightarrow f^{(n)}(x) = a^x (\ln a)^n$
Example:
Find the nth derivative of $f(x) = e^{3x}$
Solution: $f^{(n)}(x) = 3^n e^{3x}$
2. Trigonometric Functions Patterns
Sine and Cosine Functions
General Formulas:
• $f(x) = \sin(ax + b) \Rightarrow f^{(n)}(x) = a^n \sin\left(ax + b + \frac{n\pi}{2}\right)$
• $f(x) = \cos(ax + b) \Rightarrow f^{(n)}(x) = a^n \cos\left(ax + b + \frac{n\pi}{2}\right)$
💡 Memory Aid
Every derivative of sin/cos adds $\frac{\pi}{2}$ to the angle. The pattern cycles every 4 derivatives:
$f(x)$
$\sin x$
$f'(x)$
$\cos x$
$f''(x)$
$-\sin x$
$f'''(x)$
$-\cos x$
Example:
Find the 5th derivative of $f(x) = \sin(2x)$
Solution: $f^{(5)}(x) = 2^5 \sin\left(2x + \frac{5\pi}{2}\right) = 32 \sin\left(2x + \frac{\pi}{2}\right) = 32 \cos(2x)$
Note: $\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$, and $\sin(\theta + 2\pi) = \sin\theta$
3. Exponential & Logarithmic Patterns
Logarithmic Functions
General Formula:
For $f(x) = \ln(ax + b)$:
Example:
Find the 4th derivative of $f(x) = \ln(2x + 1)$
Solution: $f^{(4)}(x) = \frac{(-1)^{3} \cdot 2^4 \cdot 3!}{(2x + 1)^4} = \frac{-16 \cdot 6}{(2x + 1)^4} = \frac{-96}{(2x + 1)^4}$
4. Leibniz Rule for Product of Functions
Product of Two Functions
Leibniz Theorem:
If $y = u(x)v(x)$, then:
Where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.
💡 Application Strategy
- Choose $u(x)$ as the function whose derivatives become simpler or zero
- Choose $v(x)$ as the function whose derivatives follow a known pattern
- Look for points where higher derivatives become zero to simplify calculations
Example 1:
Find the 3rd derivative of $f(x) = x^2 e^{3x}$
Solution: Let $u(x) = x^2$, $v(x) = e^{3x}$
Using Leibniz rule:
$(uv)^{(3)} = \binom{3}{0}u^{(0)}v^{(3)} + \binom{3}{1}u^{(1)}v^{(2)} + \binom{3}{2}u^{(2)}v^{(1)} + \binom{3}{3}u^{(3)}v^{(0)}$
Since $u^{(3)}(x) = 0$, we get:
$f^{(3)}(x) = 1 \cdot x^2 \cdot 27e^{3x} + 3 \cdot 2x \cdot 9e^{3x} + 3 \cdot 2 \cdot 3e^{3x}$
$= 27x^2 e^{3x} + 54x e^{3x} + 18 e^{3x}$
$= e^{3x}(27x^2 + 54x + 18)$
Example 2 (JEE Main):
If $y = x^3 \sin x$, find $y^{(4)}$
Solution: Let $u(x) = x^3$, $v(x) = \sin x$
Using Leibniz rule for n=4:
Since $u^{(4)}(x) = u^{(5)}(x) = \cdots = 0$, only first 4 terms matter:
$y^{(4)} = \binom{4}{0}u^{(0)}v^{(4)} + \binom{4}{1}u^{(1)}v^{(3)} + \binom{4}{2}u^{(2)}v^{(2)} + \binom{4}{3}u^{(3)}v^{(1)}$
$= 1 \cdot x^3 \cdot \sin(x + 2\pi) + 4 \cdot 3x^2 \cdot \sin(x + \frac{3\pi}{2}) + 6 \cdot 6x \cdot \sin(x + \pi) + 4 \cdot 6 \cdot \sin(x + \frac{\pi}{2})$
$= x^3 \sin x + 12x^2 (-\cos x) + 36x (-\sin x) + 24 \cos x$
$= x^3 \sin x - 12x^2 \cos x - 36x \sin x + 24 \cos x$
5. Special Patterns & Shortcuts
Reciprocal Functions Pattern
Key Formula:
For $f(x) = \frac{1}{ax + b}$:
Example:
Find the 5th derivative of $f(x) = \frac{1}{2x - 1}$
Solution: $f^{(5)}(x) = \frac{(-1)^5 \cdot 2^5 \cdot 5!}{(2x - 1)^{6}} = \frac{-32 \cdot 120}{(2x - 1)^6} = \frac{-3840}{(2x - 1)^6}$
📚 Quick Reference Table
| Function | nth Derivative Formula | Key Pattern |
|---|---|---|
| $x^m$ | $\frac{m!}{(m-n)!}x^{m-n}$ | Zero when $n > m$ |
| $e^{ax}$ | $a^n e^{ax}$ | Exponential remains |
| $\sin(ax+b)$ | $a^n \sin(ax+b+\frac{n\pi}{2})$ | Phase shift by $\frac{\pi}{2}$ |
| $\cos(ax+b)$ | $a^n \cos(ax+b+\frac{n\pi}{2})$ | Phase shift by $\frac{\pi}{2}$ |
| $\ln(ax+b)$ | $\frac{(-1)^{n-1}a^n(n-1)!}{(ax+b)^n}$ | Alternating sign |
| $\frac{1}{ax+b}$ | $\frac{(-1)^n a^n n!}{(ax+b)^{n+1}}$ | Alternating sign |
🎯 Practice Problems
Test your understanding with these JEE-level problems:
1. Find the 6th derivative of $f(x) = \sin(3x)$
2. Using Leibniz rule, find the 4th derivative of $f(x) = x^4 e^{2x}$
3. If $y = x^2 \ln(1+x)$, find $y^{(3)}$ at $x=0$
💡 JEE Exam Tips
Time-Saving Strategies:
- Memorize the basic patterns for quick recognition
- For trigonometric functions, use the phase shift method
- In Leibniz rule, identify when higher derivatives become zero
- Practice mental calculation for small values of n
Common Pitfalls:
- Forgetting the alternating sign in logarithmic derivatives
- Mishandling factorial expressions in power functions
- Confusing the phase shift in trigonometric derivatives
- Missing the binomial coefficients in Leibniz rule
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