Log derivatives

1. Jun 18, 2008

projection

i need some help understanding the following:

$$y=\frac{\sqrt{x}}{2^x}$$

$$y=lnx^{{lnx}^{{lnx}^{{lnx}^{{lnx}}}}}$$

i have some idea how to do the first one. i think it goes like this:

$$y=\frac{\sqrt{x}}{2^x}$$

$$ln y=ln (2^x)-ln(\sqrt{x})$$

$$\frac{1}{y}y'=xln (2)-\frac{1}{2}ln(x)$$

$$y'=\frac{\sqrt{x}}{2^x}[xln (2)-\frac{1}{2}ln(x)]$$

am i on the right track? or should i be doing something else?

2. Jun 18, 2008

nicksauce

Looks good to me, although you flipped your negative signs in the second step.

Edit: Actually scratch that. In your third step you didn't actually differentiate anything. When you differentiate the LHS you should do the same to the RHS.

Last edited: Jun 18, 2008
3. Jun 18, 2008

projection

oh, i see.

how about the second one? i think i have to take the ln of both sides, but from there on, i am lost/

4. Jun 18, 2008

projection

is this correct then?

$$y'=\frac{\sqrt{x}}{2^x}[\frac{1}{2x}-ln2]$$

for the 2nd one, can it be written like this?

$$y=lnx^{{lnx}^{{lnx}^{{lnx}^{{lnx}}}}}$$

$$=lnx^{({lnx})^4}$$ ?

Last edited: Jun 18, 2008
5. Jun 18, 2008

dirk_mec1

Yes now the chain rule applies.

6. Jun 19, 2008

jostpuur

Do you know what you mean by

$$\textrm{ln}x^{\textrm{ln}x}?$$

Is it

$$\textrm{ln}(x^{\textrm{ln}x})$$

or

$$(\textrm{ln}x)^{\textrm{ln}x}?$$

This suggests that the both interpretations are being used at the same time.

7. Jun 19, 2008

jostpuur

Or then it suggests that a non-standard convention

$$X^{Y^Z} \underset{\textrm{BAD}}{:=} \big(X^Y\big)^Z$$

was being used.

8. Jun 19, 2008

projection

does it matter? after taking the ln of both sides, the lnx^4 will come down any way? i don't understand.....

9. Jun 20, 2008

Gib Z

He's just wondering if the exponents apply to the entire log x expression the whole time, or just the x. It does make a difference.

Jostpurr, I believe he was using "BAD" notation lol.

10. Jun 20, 2008

jostpuur

It is true that

$$A^{\textrm{ln}(x^{\textrm{ln}x})} = A^{(\textrm{ln}x)^2}$$

is the same as

$$\big(A^{\textrm{ln}x}\big)^{\textrm{ln}x} = A^{(\textrm{ln}x)^2},$$

but these are different from the

$$A^{(\textrm{ln}x)^{\textrm{ln}x}} = A^{\big((\textrm{ln}x)^{\textrm{ln}x}\big)}.$$

11. Jun 20, 2008

projection

it is written like that, there are no brackets to separate...so i don't if they want one or the other.

12. Jun 20, 2008

jostpuur

This convention

$$X^{Y^Z} = X^{\big(Y^Z\big)}$$

is standard. We can say that somebody makes a mistake if somebody puts the parentheses the other way around.

$$\textrm{ln}x^{\textrm{ln}x}$$

I'm not so sure.

These examples are standard

$$\textrm{ln}x^2 = \textrm{ln}(x^2)$$

$$\textrm{ln}^2x = (\textrm{ln}x)^2,$$

but on the other hand it would be natural to not follow this when you are writing power towers. So there seems to be two alternatives.

$$\textrm{ln}\Big(x^{\textrm{ln}\big(x^{\textrm{ln} (x^{\textrm{ln}(x^{\textrm{ln}x})})}\big)}\Big) \quad\quad\quad (1)$$

and

$$(\textrm{ln}x)^{(\textrm{ln}x)^{(\textrm{ln}x)^{(\textrm{ln}x)^{\textrm{ln}x}}}} \quad\quad\quad (2)$$

I believe (2) is right, because it would be more natural to leave the parentheses out of it, and it is not a complicated way of writing something simpler. The expression (1) is just

$$\textrm{ln}^5x$$

actually. If you cannot be sure, you can always do the both!!

Last edited: Jun 20, 2008
13. Jun 20, 2008

jostpuur

projection, you never told what you are trying to do. Is your task to find the derivatives of these functions?

14. Jun 20, 2008

projection

yes, i stated that in the title, sorry for not talking about it in the thread.

15. Jun 22, 2008

DeaconJohn

Projection,
It doesn't look to me like you were taking the derivative. It looks to me like you were taking the logarithmic derivative.
Deacon John

16. Jun 23, 2008

Gib Z

...And then solving for y'. Which is taking the derivative.

17. Jun 23, 2008

DeaconJohn

precisely.

18. Jun 23, 2008

Gib Z

...Do you not understand what was said in the last few posts? I suggest you read them again, this can be a challenge exercise: Spot the contradiction.

19. Jun 23, 2008

DeaconJohn

Gib Z,

Thanks for asking me if I understood. I think I did, but, then one can never be sure.

When I read your post ealrier post I thought you might be thinking that he was taking the derivative of "y" when in fact he was taking the derivative of "log(y)."

When I read your post above, I thought that you were not thinking wrong, that you had understoo that he was taking the derivative of log(y) and then computing the derivative of y.

Even though that's not what you said.

I was trying to give you a "kind interpretation."

Maybe I was wrong. Maybe I misundetsood you level of understanding. Maybe you missed this simple point after all.

Not to worry, all professional mathematicians make trivial mistakes like that all the time.

That is why we all try to give each other a "kind interpretatioon."

That is one of the really neat things about being a member of the community of mathematicians. Thered are other neat things too. An outstanding example is the tenderness the colleagues of John Nash treated him when he was suffering from psychophrenia. The movie "A Beuatiful Mind" does a beautiful job of presenting this aspect of the mathematical community. Too bad his wife didn't "stick it out" with him in real life like the moive has her doing.

Peace.

Deacon John

20. Jun 24, 2008

Gib Z

Well. Just in case we think these are different things, I believe taking the derivative of log y, and then solving for y' is in effect, differentiating y. Perhaps you don't, Its all good. And ill take it as a compliment you think I'm a professional mathematician =]