157k views
15 votes
Find dy/dx if y= (x²-3)³cos2x

1 Answer

7 votes

Answer:

dy/dx = 6x(x²-3)cos2x - 2(x²-3)³sin2x

Explanation:

We use the Product Rule:

y= (x²-3)³cos2x

dy/dx = (x²-3)³ d(cos2x)/dx + cos2x * d ((x²-3)³)/dx

= (x²-3)³ * -2sin2x + cos2x * 3(x²-3)*2x

= -2sin2x*(x²-3)³ + 6xcos2x*(x²-3)

= 6x(x²-3)cos2x - 2(x²-3)³sin2x

User Gopesh Gupta
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories