Asked by Arvind
d/dx 1/cos[3cosx-2sinx)/sqrt3]
Answers
Answered by
Different Steve
unbalanced parentheses, and an odd function. Is there an arccos somewhere in there?
If you meant
d/dx 1/cos[(3cosx-2sinx)/√3]
d/dx √3 sec(3cosx-2sinx)
= √3sec(3cosx-2sinx)tan(3cosx-2sinx) * (-3sinx-2cosx)
Not much that can be done with that.
If you meant
d/dx 1/cos[(3cosx-2sinx)/√3]
d/dx √3 sec(3cosx-2sinx)
= √3sec(3cosx-2sinx)tan(3cosx-2sinx) * (-3sinx-2cosx)
Not much that can be done with that.
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