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OpenStudy (anonymous):
what is the derivative of 1/sec^2 3x(tan^2 3x - 1)
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OpenStudy (anonymous):
I'm going to write tangent squared of 3x as (tan (3x))^2 The derivative will be found using the chain rule (d/dx) (tan (3x))^2 = 2(tan (3x)) (tan (3x))' = 2(tan (3x)) (sec (3x))^2 (3x)' = 2(tan (3x)) (sec (3x))^2 (3) = 6(tan (3x)) (sec (3x))^2 Asker's rating & comment
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