Derivative and Integral question

[asif zaidi]

I have two problem relating to integrals and derivatives.

Integrals: I can do but I am confused about the limit
Derivative: I am having problem. Need some help with that. I have a solution but I think it is wrong.

For each problem I have posted step by step what I am doing. Hopefully it will be easier in answering.


Problem 1: Integral

What is \int e ^{-t} (limit is from 0 (on bottom) to 1 (on top))

Problem 1: My solution
My issue is step 6

1. Take u = -t
2. Use Chain Rule: dy/dx = (dy/du)(du/dx)
3. du/dt = -1. Therefor -du = dt
4. Find limits:
For t=1: u= -1
For t=0: u= 0
5. New integral becomes (-1) \int e ^{u} du.
6. My question/issue is about the new limits:
Do I have to evaluate the integral with u=-1 on bottom of integral and u=0 on top of integral or the other way.



Problem 2 Statement:

y= x^{x^x}. Find the minimum of y.

Problem 2 Solution
I am not really sure that I am doing this right. So please advise

I know how to find the derivative of x^x.

So here are my steps

1. make a = x^x (I am taking the first two x's)
2. Therefore y = a^x
3. dy/dx = (dy/da)(da/dx)
4. Find dy/da:

y = a^x​

= e^{xlna}​

dy/da = a^x d(xlna)/da = (a^x) x^{1-x}​

5. Find da/dx

a = x^x = e^{xlnx}​

By productrule da/dx = x^x (ln x + 1)​

6. Multiple from step 4,5 and equate to 0.

My question is: is step 4 right? Actually now that I am looking at it, I am wondering if step 5 is right also. Plz advise on this.



Thanks

Asif

 

[CompuChip]

For the first problem: you have taken u = -t. Plugging in the lower boundary for t gives you the new lower boundary for u (and similarly for the upper). So the new integral runs from u = -0 below to u = -1 above. You are allowed to interchange the boundary values though, as long as you put in a minus sign, that is:
\int_a^b f(x) dx = - \int_b^a f(x) dx
In this case, it will cancel the minus you get from the substitution du -> -dt.

 

[malty]

In your first problem, the limits were from t=0 to t=1

When you substituted for u, and integrated with respect to du, you're limits were still t=o and t=1, if you were to swap these to be from t=1 to t=0 zero you would have to put a minus sign before the entire integral.
remember \int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx

So you cannot put u=-1 on the bottom without altering the sign of the integral, because you cannot put t=1 on the bottom without changin sign.

//Ah I see, compuchip already said this

For part 2, you could save yourself a lot of hassle by just letting

y=x^{x^x} =y=x^{x^2} ;)

 

[ozymandias]

Problem 1:
You should evaluate your "new" integral between the "old" limits, expressed in terms of the "new" variable u.
For example: the limit t=0 becomes u(0) = -0 = 0.
The limit t=1 becomes u(1) = -1.

Problem 2:
The chain rule works on functions of functions. In your case, you took y=a^x, while you should've taken y=x^a (i.e. the ARGUMENT of the function should be a). The final result should have more lns in it ;).

-----
Assaf
"www.physicallyincorrect.com"[/URL]

 

[asif zaidi]

I came back to this derivative question after some time. Thanks for the response on integrals

So the problem statement was to find the minimum of x^{x^x}

Problem solution

As ozymandis said I was taking the wrong approach to find derivative. Here is the proper approach.

Step1:
x^{x^x} = e^{e^{xlnx} ln x}

Step2: take derivative of above equation on right hand side

y' = x^{x^x} * derivative of (e^{e^{xlnx} ln x})

= x^{x^x} * ( (x^x /x) + ln x (x^x)' )

= x^{x^x} * ( (x^x /x) + ln x (x^x )(ln x + 1) ) --- eq1

Step3: Set eq 1 to 0 to find min

x^{x^x} * ( (x^x /x) + ln x (x^x )(ln x + 1) ) = 0

I can simplify this to

lnx (lnx + 1) + 1/x = 0 -----eq2

Question

1- How the heck do I evaluate this eq2. I ran this on MATLAB and the minimum value I get for x = 1. I never get to 0. Have I calculated the derivative properly.

2. I ran the x^x^x on MATLAB and came up with a val of 0.832 so that's the value I should come up with once I evaluate eq2 but I cannot get past above problem.

What am I doing wrong?


Thanks

Asif

 

[ozymandias]

Hi Asif,

Take q = ln(x), then solve the quadratic equation.
Edit: whoops, just saw the 1/x. :)
I don't believe this equation can be solved symbolically.
Edit 2: I just re-differentiated, your result still seems in error.
Try:
y(x) = x^{x^x} = x^a where a=x^x and use the chain rule.

Assaf
"www.physicallyincorrect.com"[/URL]

 

[asif zaidi]

Hi Ozymandias:

Seems I am doing something wrong.
Actually I tried your method of replacing a=x^x and I still got the same equation that I had before - namely the 1/x component.

Here are the steps - I cannot figure out what I am doing wrong.

Step1:
y = x^a = exp ^{alnx} where a = x^x = exp ^{xlnx}


Step2
y' = x^{x^x} (a'lnx + a/x) <--- 2nd component from chain rule. As you can see I still get the 1/x component.

Step3
a' = x^{x^x} (1+ lnx)

Step4
Substituting from step3 into step2
y' = x^{x^x} ( x^x lnx (lnx + 1) + (x^x)/x )
= (x^{x^x})(x^x) ( 1/x + lnx(lnx+1) )

Step5:To find min
0 = [/tex])(x^x) ( 1/x + lnx(lnx+1) )

Which is where I was. So obviously I am doing something wrong. Step2. I have re-read the textbook and I think I have understood it properly. I did the other problems and got the right ans.

This is driving me crazy.

Thanks

Asif