Find the derivative for the function given by Y=IN(3X-2)^4?

Question

using the general power rule

Answer 1

y = natural log (3x - 2)^4
y = 4 ln (3x - 2)
dy/dx = 4 (1/(3x-2) ) (3)
dy/dx = 12/(3x-2)

Answer 2

y = ln(3x-2)^4

This is a chain rule question, first deal with the log then the power 4 and lastly the 3x - 2, so

dy/dx = 1/((3x-2)^4) . 4(3x - 2)³ . 3

......... = 12(3x - 2)³ / ((3x-2)^4)

......... = 12 / (3x - 2)

or you can use the laws of logs

y = ln(3x-2)^4 = 4ln(3x-2) in which case still using the chain rule
dy/dx = 4. 1/(3x - 2) . 3

.......... = 12/ (3x -2) as before

Answer 3

Transform, transform, transform.
Let (3x-2)= U and dU=3dx.
Then we have Y= Ln (U)^4
Let V= U^4, and dV= 4U^3 dU (power rule)
So we have Y= Ln (V)
dY/dV * dV/dU * dU/dx = dY/dx =
... (1/V)*4(3x-2)^3*3 = 12(3x-2)^3/(3x-2)^4=
....12/(3x-2)
This is a little overblown, but it illustrates the concept of transforming.

Answer 4

(3x-2)^4