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.
3 answers
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)
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)