Is -1/2z the correct result for differentiating ln(3y-2z) with respect to z?

[cabellos]

I should know this, but i just wanted to check...differentiating ln(3y-2z) with respect to z...does this = -1/2z ?

 

[KoGs]

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)

 

[radou]

As KoGs suggested, an appropriate use of the chain rule should do just fine.

 

[cabellos]

ok so is it -2/3y + z

 

[radou]

ok so is it -2/3y + z

No, it is not. As mentioned before, try to apply the chain rule:http://mathworld.wolfram.com/ChainRule.html"

 

[cabellos]

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?

 

[radou]

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?

Now you're not going wrong, since the result is correct. Good work!