Understanding y=\frac{\sqrt{x}}{2^x} and y=lnx^{{lnx}^{{lnx}^{{lnx}^{{lnx}}}}}

[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?

 

[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.

 

[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/

 

[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?

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

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

 

[dirk_mec1]

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

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

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

Yes now the chain rule applies.

 

[jostpuur]

Do you know what you mean by

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

Is it

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

or

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

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

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

 

[jostpuur]

Or then it suggests that a non-standard convention

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

was being used.

 

[projection]

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.

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

 

[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.

 

[jostpuur]

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

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

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

 

[projection]

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

 

[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.

But about this expression

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

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!

 

[jostpuur]

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

 

[projection]

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

 

[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

 

[Gib Z]

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

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

 

[DeaconJohn]

precisely.

 

[Gib Z]

precisely.

...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.

 

[DeaconJohn]

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

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

 

[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 =]

 

[DeaconJohn]

Gib Z,

Ah, you did undeerstand after all! That was my guess. That we were just having a semantic problem.

I agree with you completely. My interest in semantic problems is minimal at best. Time to move on!

DJ