Asked by mica vega
solve for y' in the function y=sec^3(2x)-3sec2x
Answers
Answered by
Steve
Just apply the chain rule
y=sec^3(2x)-3sec2x
y' = 3 sec^2(2x) * sec(2x)tan(2x) * 2 - 3sec(2x)tan(2x)*2
= 6sec(2x)tan(2x) (sec^2(2x)-1)
= 6sec(2x)tan^3(2x)
y=sec^3(2x)-3sec2x
y' = 3 sec^2(2x) * sec(2x)tan(2x) * 2 - 3sec(2x)tan(2x)*2
= 6sec(2x)tan(2x) (sec^2(2x)-1)
= 6sec(2x)tan^3(2x)
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