Question

Is the derivative of y=2sec(3x)

= 6sec^2(3x)*3tan(3x)

Answer 1

I'll double-check, but it looks like there is too much going on there.

Answer 2

The derivative of sec(x) is sec(x)*tan(x)

Answer 3

so I'm not sure how you got the sec^2

Answer 4

\(\large 2sec(3x)=\frac{2}{cos(3x)}\)
Can use quotient rule (and a little chain rule), to get \(\large \frac{(-2)(3)(-sin(3x))}{cos^2(3x)}\)

Answer 5

I combined the two sec(3x)?

Answer 6

oh, when you derive 2*sec(x), you pull out the 2 and ignore it when you derive

Answer 7

y=2sec(3x)
(2)*(3sec(3x))+(0)*(sec(3x))
2*3sec(3x)*3tan(3x)+0*sec(3x)
6sec(3x)*3tan(3x)+sec(3x)
6sec^2(3x)*3tan(3x)


this is what i did.

Answer 8

here's what you should do

y = 2*sec(3x)

y' = d/dx[2*sec(3x)]

y' = 2*d/dx[sec(3x)]*d/dx[3x]

y' = 2*sec(3x)*tan(3x)*3

y' = 6*sec(3x)*tan(3x)

Answer 9

so i dont use the product rule?

Answer 10

no

Answer 11

2 is a constant which you can pull out

Answer 12

leaving you with you to derive sec(3x) to get sec(3x)*tan(3x) then you multiply that with the derivative of 3x (the inner function), which is 3

Answer 13

do i always do that when its multiplied by a constant?

Answer 14

yes

Answer 15

In general,

If y = k*f(x)

then

y' = k*f ' (x)

where k is a constant and f(x) is some function

Answer 16

then why did you multiply by 3x at the end

Answer 17

derivative of 3x*

Answer 18

that's from the chain rule

Answer 19

(d/dx)sec(x) = sec(x) * tan(x)

(d/dx)sec(3x) = 3 * sec(3x) * tan(3x)

(d/dx)2sec(3x) = 6 * sec(3x) * tan(3x)

Notice how adding the constant 2 doesn't effect anything. It just comes along for the ride at the end.

Answer 20

ohhh okayy.. i got it!

Answer 21

that's great

Answer 22

thanks you guys!

Answer 23

yw