How do we find the derivative of the following ...

A

Agent Smith

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[imath]y = x^{\frac{1}{x}}[/imath]?

My knowledge of differential calculus is limited to ... for [imath]y = ax^n[/imath], [imath]\frac{dy}{dx} = nax^{n - 1}[/imath]
 
Agent Smith said:
[imath]y = x^{\frac{1}{x}}[/imath]?

My knowledge of differential calculus is limited to ... for [imath]y = ax^n[/imath], [imath]\frac{dy}{dx} = nax^{n - 1}[/imath]
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What about [imath]\displaystyle \ln y = \frac{1}{x}\ln x[/imath] and use implicity differentiation?
 
So If [imath]u = \ln y = \frac{1}{x} \ln x[/imath] then I know that [imath]\frac{du}{dx} = \frac{du}{dy} \times \frac{dy}{dx}[/imath]
[imath]u[/imath] is a function of [imath]y[/imath] is a function of [imath]x[/imath]

This is from last night's reading session.
 
The suggestion was to use logarithmic differentiation; look it up in your book. Unfortunately, it is probably considerably later than where you are.

Please show us an image of the problem, and the title of the chapter, so we can confirm what you might be expected to do. This can't really be done without the more advanced methods, so I suspect they are asking for something other than what you think..
 
Dr.Peterson said:
The suggestion was to use logarithmic differentiation; look it up in your book. Unfortunately, it is probably considerably later than where you are.

Please show us an image of the problem, and the title of the chapter, so we can confirm what you might be expected to do. This can't really be done without the more advanced methods, so I suspect they are asking for something other than what you think..
Click to expand...
I doubt it if you'll be interested in the original problem. I was exploring sets and I found out that for a set A, whose cardinality n(A) = p, the power set of the set A, P(A) has a cardinality n(P(A)) = [imath]2^p[/imath]
Then this: What if [imath]p = 2^p[/imath]? Fro that [imath]p^{\frac{1}{p}} = 2[/imath]. The general form is [imath]y = x^{\frac{1}{x}}[/imath]
I graphed the function on Desmos and discovered that it has a maximum at [imath](e, 1.4446...)[/imath] where [imath]e[/imath] is Euler's number. That is to say [imath]e^{\frac{1}{e}} = 1.4446...[/imath], I wanted to confirm this using maxima & minima differentiation.
 
Agent Smith said:
So If [imath]u = \ln y = \frac{1}{x} \ln x[/imath] then I know that [imath]\frac{du}{dx} = \frac{du}{dy} \times \frac{dy}{dx}[/imath]
[imath]u[/imath] is a function of [imath]y[/imath] is a function of [imath]x[/imath]

This is from last night's reading session.
Click to expand...
You do not have to do it like this. You can do it directly.

[imath]\displaystyle \frac{d}{dx}\left(\ln y = \frac{1}{x}\ln x\right)[/imath]


[imath]\displaystyle \frac{d}{dx}\left(\ln y\right) = \frac{d}{dx}\left(\frac{1}{x}\ln x\right)[/imath]


[imath]\displaystyle \frac{d}{dy}\left(\ln y\right)\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{x}\ln x\right)[/imath]
 
So for [imath]f(x) = x^{\frac{1}{x}}[/imath], we have [imath]f'(x)[/imath]. At maxima, [imath]f'(x) = 0[/imath] and [imath]x = e[/imath]? 🤔
 
Agent Smith said:
I doubt it if you'll be interested in the original problem. I was exploring sets
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What you're saying is that this is not a problem from the book you say you're reading. That explains why it would require more than you've learned yet. That's all we need to know about the origin of your question.

Agent Smith said:
So for [imath]f(x) = x^{\frac{1}{x}}[/imath], we have [imath]f'(x)[/imath]. At maxima, [imath]f'(x) = 0[/imath] and [imath]x = e[/imath]? 🤔
Click to expand...
Yes, that's what you'll get.

In order to complete the work, you'll need to know the product rule as well as the chain rule. (And by the way, it's entirely possible that either your question, or [imath]x^x[/imath], will be an example in your book's discussion of logarithmic differentiation.)
 
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