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If y=cox^cosx^cosx^infinty then dy/dx is equal to -y^2tanx/1-ylncosx y^2tanx/1+ylncosx

If y=cox^cosx^cosx^infinty then dy/dx is equal to
 
  1. -y^2tanx/1-ylncosx
  2. y^2tanx/1+ylncosx

Grade:12

1 Answers

Arun
25758 Points
4 years ago
Dear Anirudh
 
y = cos x^ cos x ^ cosx^ .... infinity
 
y = (cosx)^y
 
Now take log both the sides and then differentiate
 
You will get the desired result

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