NCERT Class 12 · Ch 5 Exercise 5.7 · Q7
NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.7 Question 7
Find the second derivative of \(e^{6 x} \cos{\left(3 x \right)}\) with respect to x; on R.
💡 Concept Hints
- Identify the domain and the inner and outer functions, or the relevant one-sided limits.
- Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: \(9 \left(- 4 \sin{\left(3 x \right)} + 3 \cos{\left(3 x \right)}\right) e^{6 x}\)
- Work on R.
- Product rule: differentiate each variable factor once, keeping the other factors unchanged. Here \(\frac{d}{d x} e^{6 x} \cos{\left(3 x \right)}\) = \(- 3 e^{6 x} \sin{\left(3 x \right)} + 6 e^{6 x} \cos{\left(3 x \right)}\).
- Differentiate the first derivative \(- 3 e^{6 x} \sin{\left(3 x \right)} + 6 e^{6 x} \cos{\left(3 x \right)}\) again.
- Product rule: differentiate each variable factor once, keeping the other factors unchanged. Here \(\frac{d}{d x} \left(- 3 e^{6 x} \sin{\left(3 x \right)}\right)\) = \(- 18 e^{6 x} \sin{\left(3 x \right)} - 9 e^{6 x} \cos{\left(3 x \right)}\).
- Product rule: differentiate each variable factor once, keeping the other factors unchanged. Here \(\frac{d}{d x} 6 e^{6 x} \cos{\left(3 x \right)}\) = \(- 18 e^{6 x} \sin{\left(3 x \right)} + 36 e^{6 x} \cos{\left(3 x \right)}\).
- Sum rule: \(\frac{d}{d x} \left(- 3 e^{6 x} \sin{\left(3 x \right)} + 6 e^{6 x} \cos{\left(3 x \right)}\right)\) = \(- 36 e^{6 x} \sin{\left(3 x \right)} + 27 e^{6 x} \cos{\left(3 x \right)}\).
- Combining the terms gives y″ = \(9 \left(- 4 \sin{\left(3 x \right)} + 3 \cos{\left(3 x \right)}\right) e^{6 x}\).
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