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Mathematics 18 Online
OpenStudy (anonymous):

what is the derivative of ln(x/2)?

OpenStudy (tkhunny):

What is \(\dfrac{d}{dx}ln(x)\)?

OpenStudy (anonymous):

1/x

OpenStudy (anonymous):

does it help to know that \[\ln(\frac{x}{2})=\ln(x)-\ln(2)\]?

OpenStudy (anonymous):

not really?

OpenStudy (tkhunny):

Okay, now remember the Chain Rule and solve your problem. Or, if you wish, the logarithm transformation can be helpful. I'd do it both ways, just to make sure it worked.

OpenStudy (tkhunny):

What's \(\dfrac{d}{dx}ln(2)\)? (The derivative of a constant?)

OpenStudy (anonymous):

zero

OpenStudy (tkhunny):

Okay, then what is \(\dfrac{d}{dx}\left(ln(x) - ln(2)\right)\)?

OpenStudy (anonymous):

1/x - 0

OpenStudy (anonymous):

is that right?

OpenStudy (tkhunny):

Well, the "- 0" is a little silly. Now do it the other way. What is \(\dfrac{d}{dx}ln\left(\dfrac{x}{2}\right)\)? Don't forget the Chain Rule.

OpenStudy (anonymous):

1/x

OpenStudy (anonymous):

is that right?

OpenStudy (tkhunny):

I hope so, since that what we managed the other way. How did you get that?

OpenStudy (anonymous):

the way you showed me

OpenStudy (tkhunny):

You didn't use the Chain Rule. You used the log rules. Can you do it WITHOUT the log rules?

OpenStudy (anonymous):

no i haven't taken calculus since high school...

OpenStudy (tkhunny):

We, then let's just go with that as it stands. Just for exposure, it looks like this: \(\dfrac{d}{dx}ln\left(\dfrac{x}{2}\right) = \dfrac{1}{x/2}\cdot (1/2) = \dfrac{1}{x}\) Same result. You will need the Chain Rule. Keep your eyes out for it when you trip over it again. Good work!

OpenStudy (anonymous):

Thanks!

OpenStudy (ivancsc1996):

I think the answer is wrong. Just take aut the constant and differentiate:\[\frac{ d }{ dx }\ln \frac{ x }{ 2 }=\frac{ 1 }{2 }\frac{ d }{ dx }\ln x=\frac{ 1 }{2 }\frac{ 1 }{ x }=\frac{ 1 }{ 2x }\]

OpenStudy (ivancsc1996):

Oh no, I just saw what i did wrong. It is 1/x.

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