Natural Log in a Derivative: What Are the Rules?

[MacLaddy]

Homework Statement

I have a problem that I'm working on that I have almost solved, yet I am just a tad off of what the book says the answer is. I will show the way I'm doing it, and where I depart from the steps the book takes.

The graph of $y=(x^2)^x$ has two horizontal tangent lines. Find equations for both of them. (I bolded "two" because this is part of my mistake)

Homework Equations

Chain rule, exponential rule, etc

The Attempt at a Solution

$y=(x^2)^x = x^{2x}$

$\ln{y}=\ln{x^{2x}}$

This is where the book and I deviate. This is what I do.

$lny = 2xlnx$

Taking the derivative

$\frac{1}{y}y'=2x\frac{1}{x}+2lnx$

$\frac{1}{y}y'=2+2lnx$

$y'=y(2+2lnx)$

$y'=x^{2x}(2+2lnx)$

Solving for zero

$x^{2x} = 0$ when$x=0$

$2+2lnx = 0$

$2lnx = -2$

$lnx = \frac{-2}{2}$

$lnx = -1$

$e^{lnx} = e^{-1}$

$x = e^{-1} = 1/e$

Here is where the problem lies, the book never brought the exponent in front, like $lny = 2xlnx$. Instead it kept $lny = lnx^{2x}$ and then found the derivative, which ends up with two answers.

What are the rules here? Any advice (about this problem) is appreciated.

Mac

 

[Ray Vickson]

Homework Statement

I have a problem that I'm working on that I have almost solved, yet I am just a tad off of what the book says the answer is. I will show the way I'm doing it, and where I depart from the steps the book takes.

The graph of $y=(x^2)^x$ has two horizontal tangent lines. Find equations for both of them. (I bolded "two" because this is part of my mistake)

Homework Equations

Chain rule, exponential rule, etc

The Attempt at a Solution

$y=(x^2)^x = x^{2x}$

$\ln{y}=\ln{x^{2x}}$

This is where the book and I derivate (no pun intended) This is what I do.

$lny = 2xlnx$

Taking the derivative

$\frac{1}{y}y'=2x\frac{1}{x}+2lnx$

$\frac{1}{y}y'=2+2lnx$

$y'=y(2+2lnx)$

$y'=x^{2x}(2+2lnx)$

Solving for zero

$x^{2x} = 0$ when$x=0$

$2+2lnx = 0$

$2lnx = -2$

$lnx = \frac{-2}{2}$

$lnx = -1$

$e^{lnx} = e^{-1}$

$x = e^{-1} = 1/e$

Here is where the problem lies, the book never brought the exponent in front, like $lny = 2xlnx$. Instead it kept $lny = lnx^{2x}$ and then found the derivative, which ends up with two answers.

What are the rules here? Any advice (about this problem) is appreciated.

Mac

The function (x^2)^x is well-defined for x < 0 or x > 0, but the function exp(2*x*ln(x)) is not defined for x < 0: you would be taking the log of a negative number. The two expressions are the same only if x > 0.

RGV

 

[MacLaddy]

Thanks for the fast reply, Ray, but I am not sure I understand.

Are you saying that because $x^{2x}$ is not defined as greater than zero, then you can not change $lnx^{2x}$ to $2xlnx$?

If that's true, then why does that exponent rule apply wrt Ln? I think I'm still missing something.

 

[SammyS]

Homework Statement

I have a problem that I'm working on that I have almost solved, yet I am just a tad off of what the book says the answer is. I will show the way I'm doing it, and where I depart from the steps the book takes.

The graph of $y=(x^2)^x$ has two horizontal tangent lines. Find equations for both of them. (I bolded "two" because this is part of my mistake)

Homework Equations

Chain rule, exponential rule, etc

The Attempt at a Solution

$y=(x^2)^x = x^{2x}$

$\ln{y}=\ln{x^{2x}}$

This is where the book and I derivate (no pun intended) This is what I do.   (The word is deviate.)

$lny = 2xlnx$

Taking the derivative

$\frac{1}{y}y'=2x\frac{1}{x}+2lnx$

$\frac{1}{y}y'=2+2lnx$

$y'=y(2+2lnx)$

$y'=x^{2x}(2+2lnx)$

Solving for zero

$x^{2x} = 0$ when$x=0$

$2+2lnx = 0$

$2lnx = -2$

$lnx = \frac{-2}{2}$

$lnx = -1$

$e^{lnx} = e^{-1}$

$x = e^{-1} = 1/e$

Here is where the problem lies, the book never brought the exponent in front, like $lny = 2xlnx$. Instead it kept $lny = lnx^{2x}$ and then found the derivative, which ends up with two answers.

What are the rules here? Any advice (about this problem) is appreciated.

Mac

First of all, $(x^2)^x = x^{2x}$ only for x ≥ 0.

More generally you can write, $(x^2)^x = |x|^{2x}\,,$ which is true for all real values of x.

Therefore, $\displaystyle \ln(y)=\left\{ \begin{array}{rcl} 2x\ln(x)& \text{ if } & x\ge 0 \\ \\ 2x\ln(-x)& \text{ if } & x<0\end{array}\right.$

Work with that.

 

[MacLaddy]

Sorry guys, but this isn't making any sense to me. Is there a reference you can point me to?

I understand that for ln(x), x > 0, but I'm not seeing how that applies to the exponent, and why in your piecewise function above you can put a negative in the Natural Log. I also do not understand why x for $x^{2x}$ must be equal to or greater than zero.

The rule as I've always seen it is $\ln{a^b} = b*lna$

 

[SammyS]

*EDIT*

Jeopardy music playing in my head. It's about to click... but not yet.

Right.

If x = -3/2, then $\displaystyle (x^2)^x = \left(\left(\frac{-3}{2}\right)^2\right)^{-3/2}=\left(\frac{9}{4}\right)^{-3/2}=\frac{8}{27}\,,$ whereas, $\displaystyle x^{2x} = \left(\frac{-3}{2}\right)^{2(-3/2)}=\left(\frac{-3}{2}\right)^{-3}=\frac{-8}{27}\,,$

 

[HallsofIvy]

Sorry guys, but this isn't making any sense to me. Is there a reference you can point me to?

I understand that for ln(x), x > 0, but I'm not seeing how that applies to the exponent, and why in your piecewise function above you can put a negative in the Natural Log. I also do not understand why x for $x^{2x}$ must be equal to or greater than zero.

The rule as I've always seen it is $\ln{a^b} = b*lna$

Then you need to go back an review. What you should have seen was $\ln{a^b}= b\ln{a}$ if a> 0[/itex]. You keep ignoring the possibility that a is negative, in which case you cannot take the derivative.

 

[SammyS]

If x itself is a negative number, as in, x = -5, then ln(-x) = ln(-(-5)) = ln(5).

That's how you can have ln(-x) .

 

[MacLaddy]

Right.

If x = -3/2, then $\displaystyle (x^2)^x = \left(\left(\frac{-3}{2}\right)^2\right)^{-3/2}=\left(\frac{9}{4}\right)^{-3/2}=\frac{8}{27}\,,$ whereas, $\displaystyle x^{2x} = \left(\frac{-3}{2}\right)^{2(-3/2)}=\left(\frac{-3}{2}\right)^{-3}=\frac{-8}{27}\,,$

Ah, I see. I was not aware of this rule.

Then you need to go back an review. What you should have seen was $\ln{a^b}= b\ln{a}$ if a> 0[/itex]. You keep ignoring the possibility that a is negative, in which case you cannot take the derivative.

Yes, definitely need to review. I just want to make sure I know exactly what to review.

If x itself is a negative number, as in, x = -5, then ln(-x) = ln(-(-5)) = ln(5).

That's how you can have ln(-x) .

That makes more sense, thanks for the clarification.

I am going to dig into this question for a bit, and hopefully I will understand it. I think I have some serious deficiencies in my understanding somewhere. I'm just not sure if it goes back to algebraic rules I should know, or if this is something new in calculus that I'm not grasping.

I appreciate both of your help. If I do have more questions, would you recommend starting a new thread, or continuing with this one?