MathDerivatives.
posted by Anonymous .
if y=sec^(3)2x, then dy/dx=
my answer is 6 sec^3 2xtan 2x but im not sure if that is right. ?
Can someone show me if i did it correct.

good job

Sorry to bother you, can you explain the steps to me, so I can see the work and check my work. I thought i had done it wrong, I started my work and then guessed.

write it as
y = (sec (2x))^3
now use the chain rule ...
y' = 3(sec (2x)^2 (derivative of sec(2x))
= 3(sec (2x)^2(sec (2x))tan (2x))(2)
= 6(sec (2x))^3 tan (2x)
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