Finding dy/dx for y = 4ln3x using the chain rule

[james03]

Would like help with this question

Question:y = 4ln3x, find dy/dx

Homework Equations

1. I used the chain rule:

y=4ln u and u =3x

dy/du=4/u THIS IS MY PROBLEM

There is no number between the ln and u, eg. ln4u, so WHY do you use
d/dx ln(ax+b)=a/(ax+b)?

The Attempt at a Solution

y=4ln u and u =3x

dy/du=4/u du/dx=3

dy/dx=12/3x=4x.

 

[Millennial]

If I got it right, you mean $\displaystyle \frac{d}{dx}4\log(3x) = 4\frac{d}{dx}\log(3x)$.

Now, it should be already plain obvious to some people what the answer to this is, but in case it isn't, there are two ways to go here. One is to use the fundamental identity of logarithms (how does a logarithm distribute over a product?), and the other is to use the chain rule. Pick whichever you want.

 

[jedishrfu]

your last eqn: dy/dx=12/3x=4x should be = 4/x right?

 

[Mark44]

Would like help with this question

Question:y = 4ln3x, find dy/dx

Homework Equations

1. I used the chain rule:

y=4ln u and u =3x

dy/du=4/u THIS IS MY PROBLEM

There is no number between the ln and u, eg. ln4u, so WHY do you use
d/dx ln(ax+b)=a/(ax+b)?

The Attempt at a Solution

y=4ln u and u =3x

dy/du=4/u du/dx=3

dy/dx=12/3x=4x.

dy/du = 4/u
dy/dx = dy/du * du/dx = 4/u * 3 = ?

 

[SammyS]

You might also notice that $4\ln(3x)=4(\ln(3)+\ln(x))$

$=4\ln(3)+4\ln(x)\ .$​