Calculating Derivatives of a Function on an Open Interval ]0,1[

[danni7070]

Homework Statement

Lets define a function f on an open interval ]0,1[ by setting

[tex] f(x) = \int_{1/2}^\sqrt{x}} tln\frac{1+t^2}{1-t^2}dt [/tex]

Calculate the first three deriatives f'(x), f''(x) and f'''(x)

Homework Equations

Product and Quotent rule.

The Attempt at a Solution

Ok. First of all. Does that open interval ]0,1[ have any meaning to the calculations?

I took [tex] f'(x) = \sqrt{x} ln \frac{1+x}{1-x}dx - \frac{1}{2} ln \frac{1+\sqrt{1/2}}{1-\sqrt{1/2}} [/tex]

Is that the right way to do it ?

 

[Dick]

That's right. The integrand is undefined at 1 and 0. Hence the restriction to (0,1).

 

[Dick]

Ooops. Correction. Your constant part shouldn't be there. If the antiderivative of f(t) is F(t), the integral from a to x is F(x)-F(a). The derivative of that is just F'(x)=f(x). The a part disappears.

 

[danni7070]

Jesus Christ. Of course. x cannot be 0 because of sqrt(x) and not 1 because of the (1-x) thing.

I just thougt that it was ment to trick you to say "lets define on an open interval..."

Clearly it did in my case.

Thanks Dick.

[tex] f ' (x) = \sqrt{x}ln\frac{1+x}{1-x} - CONSTANT[/tex]

d/dx constant = 0


[tex] d/dx \sqrt{x} = \frac {1}{2\sqrt{x}} [/tex]


[tex] u = \frac{1+x}{1-x} [/tex] and [tex] u ' = \frac{(1-x)+(1+x)}{(1-x)^2} = \frac {2}{(1-x)^2} [/tex]

[tex] g(u) = ln(u) [/tex] and [tex] g ' (u) = \frac{1}{x} [/tex]

[tex] d/dx ln\frac{1+x}{1-x} = \frac{1}{x} \frac{2}{(x-1)^2} = \frac{2}{x(1-x)^2} [/tex]

so...


[tex] f '' (x) = \frac {ln\frac{1+x}{1-x}}{2\sqrt{x}} + \frac{2\sqrt{x}}{x(1-x)^2} [/tex]

Is this correct? And if it is, can I simplify the result ?

 

[Dick]

As I explained in my last note drop the constant in f'(x). f(x) COULD be defined on [0,1), but they chose not to (it is defined at 0 - I was thinking they had a log(x) in there somewhere). I'm not getting the same thing as you when I differentiate the log. But you might find it easier to handle the log by writing log((1+x)/(1-x))=log(1+x)-log(1-x).

 

[danni7070]

Jeez if i do that with the log I will be forever doing this. Or maybe not.

EDIT:

[tex] f ' (x) = \sqrt{x}ln(1+x) - \sqrt{x}ln(1-x) [/tex]

[tex] d/dx \sqrt{x} ln(1+x) = \frac{ln(1+x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1+x}[/tex]

[tex] d/dx \sqrt{x}ln(1-x) = \frac{ln(1-x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1-x} [/tex]

so

[tex] f '' (x) = (\frac{ln(1+x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1+x}) - (\frac{ln(1-x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1-x}) [/tex]

and for f'''(x) I just go quatant crazy.

 

[danni7070]

[tex] f ''' (x) = \frac{\frac{2\sqrt{x}}{(1+x)} - \frac{ln(1+x)}{\sqrt{x}}}{4x} [/tex]

Correct? No, its incorrect...

 

[Dick]

You are forgetting a parenthesis in the FIRST derivative - both logs are multiplied by sqrt(x). But I just meant to differentiate the log by differentiating log(1+x)-log(1-x) -> 1/(1+x)+1/(1-x) -> 2/(1-x^2).

 

[danni7070]

[tex] f ' (x) = \sqrt{x}ln(1+x) - \sqrt{x}ln(1-x) [/tex]

[tex] d/dx \sqrt{x} ln(1+x) = \frac{ln(1+x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1+x}[/tex]

[tex] d/dx \sqrt{x}ln(1-x) = \frac{ln(1-x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1-x} [/tex]

so

[tex] f '' (x) = (\frac{ln(1+x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1+x}) - (\frac{ln(1-x)}{2\sqrt{x}} + \frac{\sqrt{x}}{1-x}) [/tex]

 

[Dick]

You have a sign error on the last term, the derivative of log(1-x)=(-1)/(1-x). One more derivative to go.

 

[danni7070]

Ok, I think I got it now. And yes of course there is a minus sign :)

I just put the result I got

[tex] f ''' (x) = (\frac{\frac{2\sqrt{x}}{1+x}-\frac{ln(1+x)}{\sqrt{x}}}{4x} + \frac{\frac{1+x}{2\sqrt{x}}-\sqrt{x}}{(1+x)^2})-(\frac{\frac{2\sqrt{x}}{1-x}+\frac{ln(1-x}{\sqrt{x}}}{4x} - \frac{\frac{1-x}{2\sqrt{x}} + \sqrt{x}}{(1-x)^2})[/tex]

 

[Dick]

Hope you're right. It's kind of hopeless to check something like that. Even if you find something, it's just as likely to be a TeX error as anything else. I know you basically know what you are doing.

 

[danni7070]

Yes, but thanks a lot for the help. But what's the meaning of finding f'''(x)

f'(x) tells you where the CP's are ans so..

f''(x) tellst you where the lines turn

but what the hell does f'''(x) do other then robbing you valiable time?

 

[Dick]

It tells you where the second derivative is increasing and decreasing. Which is not something you generally need when sketching a graph. Here it's just for practice (i.e. robbing you of valuable time).