- #1
cabellos
- 76
- 1
I should know this, but i just wanted to check...differentiating ln(3y-2z) with respect to z...does this = -1/2z ?
Is -1/2z the correct result for differentiating ln(3y-2z) with respect to z?
- Thread starter cabellos
- Start date
-
- Tags
- Differentiation
Click For Summary
Homework Help Overview
The discussion revolves around differentiating the function ln(3y-2z) with respect to the variable z. Participants are exploring the correct application of differentiation techniques, particularly the chain rule, in this context.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants are questioning the correctness of the initial differentiation result of -1/2z and discussing the proper application of the chain rule. There are attempts to clarify the steps involved in differentiation and to verify the results through integration.
Discussion Status
The discussion is active with participants providing feedback on each other's reasoning. Some guidance has been offered regarding the application of the chain rule, and there is an acknowledgment of a correct result in one of the posts. However, there is no explicit consensus on the final answer yet.
Contextual Notes
Participants are navigating through the differentiation process and checking assumptions about the function and its derivatives. There is a focus on ensuring that the differentiation is applied correctly, particularly in relation to the chain rule.
- #2
KoGs
- 106
- 0
You must differentiate the whole thing first, then differentiate what is on the inside.
To check your answer take the integral of -1/2z, you will see it is not equal ln(3y-2z)
To check your answer take the integral of -1/2z, you will see it is not equal ln(3y-2z)
- #3
radou
Homework Helper
- 3,149
- 8
As KoGs suggested, an appropriate use of the chain rule should do just fine.
- #4
cabellos
- 76
- 1
ok so is it -2/3y + z
- #5
radou
Homework Helper
- 3,149
- 8
cabellos said:
No, it is not. As mentioned before, try to apply the chain rule:http://mathworld.wolfram.com/ChainRule.html"
Last edited by a moderator:
- #6
cabellos
- 76
- 1
I did apply it...this is how i calculated that result:
d/dz In(3y-2z)
y=In u therefore dy/du = 1/u
u=3y-2z therefore du/dz = -2
dy/du x du/dz = -2/(3y-2z)
where am i going wrong?
d/dz In(3y-2z)
y=In u therefore dy/du = 1/u
u=3y-2z therefore du/dz = -2
dy/du x du/dz = -2/(3y-2z)
where am i going wrong?
- #7
radou
Homework Helper
- 3,149
- 8
cabellos said:
Now you're not going wrong, since the result is correct. Good work!
Similar threads
- · Replies 6 ·
- · Replies 13 ·
- · Replies 5 ·
- · Replies 3 ·