Derivative of Natural Log

**Chris22** · October 28th 2009, 02:24 PM

Hey, I was just wondering if anyone could walk me through how to find the derivative of $y^2=2x ln(3x)$

I think that the beginning of the solution should look something like this $(2y)y'= 2x(\frac{1}{3x})+2 ln(3x)$ , but I am not sure.

Thank you for your help.

**e^(i*pi)** · October 28th 2009, 02:35 PM

Originally Posted by Chris22

Hey, I was just wondering if anyone could walk me through how to find the derivative of $y^2=2x ln(3x)$

I think that the beginning of the solution should look something like this $(2y)y'= 2x(\frac{1}{3x})+2 ln(3x)$ , but I am not sure.

Thank you for your help.

Recall that $\frac{d}{dx}(ln[f(x)]) = \frac{f'(x)}{f(x)}$

**WhoCares357** · October 28th 2009, 02:36 PM

Originally Posted by Chris22

Hey, I was just wondering if anyone could walk me through how to find the derivative of $y^2=2x ln(3x)$

I think that the beginning of the solution should look something like this $(2y)y'= 2x(\frac{1}{3x})+2 ln(3x)$ , but I am not sure.

Thank you for your help.

All that's left now is to simplify (if you are required to do so).

Just move 2y to the other side and substitute y with what it equals.

In your case $y=\sqrt{2xln(2x)}$

Edit: Yeah, above post is correct. You forgot to take the derivative of the inside of ln. Otherwise it's good.

**Chris22** · October 28th 2009, 02:45 PM

Thanks for the help, but I just looked at the answers my teacher gave me and the answer it says there is $\frac{1+ln(3x)}{y}$ .
I don't know how he got that answer. Do either of you know what I am doing wrong?

Thanks again for your guys' help.

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