A conceptual problem about symmetry

[lucas.salien]

 

[Dr.Peterson]

It will help if you tell us why you think the derivative of ln(x) should be ln(x). What theorem do you think implies this? Have you tried graphing ln(x) to see what its slope looks like? Have you seen how the derivatives of other inverse functions compare to those of the original functions?

Without that to discuss, all I can say is, that just isn't the way it works! You just have some unidentified misconception.

 

[lucas.salien]

Indeed,
But because an inverse function has a symmetry by the bissectrice, what happens with the derivatifs? Is there also a symmetry?

 

[Steven G]

I guess not, since the derivative of ln(x) is not ln(x)

 

[Harry_the_cat]

You have said that the integral of e^x wrt x is e^x. Shouldn't it be e^x + c.
Maybe that has something to do with it.

 

[Dr.Peterson]

But because an inverse function has a symmetry by the bissectrice, what happens with the derivatifs? Is there also a symmetry?

I suppose what you mean by "symmetry by the bissectrice" is that the inverse function is the reflection of the original function in the line y=x that bisects the angle between the axes. Have you tried experimenting with this to see what implication it has?

What we can do to relate the derivatives of a function and its inverse is this:

In the case [imath]f(x) = e^x[/imath], this means that the derivative of [imath]f^{-1}(x) = \ln(x)[/imath] is [imath]\frac{1}{e^{\ln(x)}} = \frac{1}{x}[/imath]. That is not [imath]\ln(x)[/imath].

Take the link to see more. The main difficulty in the way of the sort of symmetry you are looking for may be that the inverse function has a different argument than the original function.

 

[pka]

The following is taken from the calculus textbook by Gillman & McDowell.
The logarithm is defined as [imath]\mathcal{L}(\alpha\beta)=\mathcal{L}(\alpha)+\mathcal{L}(\beta)\text{ where }\alpha>0~\&~\beta>0)[/imath]
If we define the [imath]\log(x)=\displaystyle \int_1^x {\frac{1}{t}dt} \text{ for all }x>o[/imath] then it easily follows that that is the same as [imath]\mathcal{L}(x)[/imath]
Moreover, it also can be shown that [imath]{D_x}\left( {\log (x)} \right) =\displaystyle \frac{1}{x}~\&~ \log (\alpha \beta ) = \log (\alpha ) + \log (\beta) [/imath]
Building on those ideas we then define [imath]\bf e[/imath] as the number that [imath]\log({\bf e})=\displaystyle\int_1^{\bf{\large e}} {\frac{1}{t}dt} = 1[/imath]
The we expand to get [imath]y=\exp(x)\;\:[/imath] means [imath]\;\;x=log(y)[/imath] and because [imath]\log(e)=1[/imath] we have [imath]\exp(1)=e[/imath]

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[lucas.salien]

I suppose what you mean by "symmetry by the bissectrice" is that the inverse function is the reflection of the original function in the line y=x that bisects the angle between the axes. Have you tried experimenting with this to see what implication it has?

What we can do to relate the derivatives of a function and its inverse is this:

In the case [imath]f(x) = e^x[/imath], this means that the derivative of [imath]f^{-1}(x) = \ln(x)[/imath] is [imath]\frac{1}{e^{\ln(x)}} = \frac{1}{x}[/imath]. That is not [imath]\ln(x)[/imath].

Take the link to see more. The main difficulty in the way of the sort of symmetry you are looking for may be that the inverse function has a different argument than the original function.

Thanks,
I can see clearly now.....