Question

find the derivative f(x)=ln(3x-2)^8

Answer 1

is that

[ln(3x-2)]^8

or is it

ln[ (3x-2)^8 ]

Answer 2

\[f(x)=\ln(3x-2)^{8}\]

Answer 3

hmm I'll go with the second one, but unfortunately what they gave you is a bit vague

Answer 4

y = ln[ (3x-2)^8 ]

can be written as

y = 8*ln[ (3x-2) ]

or just

y = 8*ln(3x-2)

Answer 5

so let's derive y = 8*ln(3x-2)

we pull the constant out and ignore it

then we derive ln(3x-2) to get 1/(3x-2)

but remember you have to use the change rule, so you're multiplying that by the derivative of what's inside: which is 3x-2....so derive 3x-2 to get 3 and multiply it with 1/(3x-2) to get 3/(3x-2)

Answer 6

so y = 8*ln(3x-2) turns into y ' = 8*(3/(3x-2))

then you just simplify to get y ' = 24/(3x-2)

Answer 7

can you explain why you can pull the 8 out front please?

Answer 8

because deriving something like

y = 10*ln(x)

or

y = 20*ln(x)

involves the same steps: pull out the constant, derive ln(x) to get 1/x, then reintroduce the constant back in

Answer 9

it's based on the rule that if y = c*f(x), then y ' = c * f ' (x)

where c is any constant

Answer 10

Alright, this derivative stuff drives me crazy

Answer 11

i understand that, but why can you pull the 8 from the exponent in front to make it a product?

Answer 12

yeah it'll derive any one crazy...

Answer 13

oh I'm using the rule that ln(x^y) = y*ln(x)

Answer 14

that's how I went from y = ln[ (3x-2)^8 ] to y = 8*ln(3x-2)

Answer 15

ahhh cool thanks!

Answer 16

np

Answer 17

thanks lol

Answer 18

no thank you

Answer 19

I'm glad it's making sense now

Answer 20

I see, so you'll have to somehow connect the dots in your head so you can remember it all, but I'm sure you can definitely do that