How do you differentiate x^{1/x} using the power rule and chain rule?

[madmike159]

I was looking at results of different numbers in the equation $$\sqrt[x]{x}$$ and found out that the biggest result came when it was $$\sqrt[e]{e}$$. I know this can be re-written as $$x^{1/x}$$ and that the gradient would be 0 at x = e. How would you differentiate y = $$x^{1/x}$$, I can't seem to do it using any of the laws I know.

 

[D H]

Find a function $f(x)$ that let's you rewrite $y=x^{1/x}$ in the form $y=\exp(f(x))$.

 

[madmike159]

I don't see how that works but I'll give it a go.

 

[lurflurf]

use the power rule
[u^v]'=v*[u^(v-1)]*u'+[u^v]*[log(u)]*v'

 

[D H]

Rather than committing the power rule to memory, I find it much easier to remember a couple simple facts which happen to be of use in a lot of other places,

1. $$\frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}\,f'(x)$$
2. $$f(x) = e^{\ln f(x)}\;\Rightarrow\;u(x)^{v(x)} = e^{v(x)\,\ln u(x)}$$

and thus

$$\aligned \frac{d}{dx}\left(u(x)^{v(x)}\right) &= \frac{d}{dx}\left(e^{v(x)\,\ln u(x)}\right) \\ &= e^{v(x)\,\ln u(x)}\frac d {dx}\left(v(x)\,\ln u(x)\right) \\ &= u(x)^{v(x)}\left(v'(x)\,\ln u(x) + \frac{v(x)\,u'(x)}{u(x)}\right)$$

 

[HallsofIvy]

Find a function $f(x)$ that let's you rewrite $y=x^{1/x}$ in the form $y=\exp(f(x))$.

Equivalently, if y= x^1/x, then ln(y)= (1/x)ln(x). Use "implicit differentiation".

 

[lurflurf]

Equivalently, if y= x^1/x, then ln(y)= (1/x)ln(x). Use "implicit differentiation".

or just use it on
y^x=x

 

[lurflurf]

Rather than committing the power rule to memory, I find it much easier to remember a couple simple facts which happen to be of use in a lot of other places,

How is a less general rule more useful?

The power rule is easy to recall since it follows from other rules
first remember the right hand side
[u^v]'
next use the chain rule (C is a constant)
[u^v]'=[u^C]'|C=v+[C^v]'|C=u
now recall
[x^C]'=C*[x^(C-1)]
or using the chain rule
[u^C]'=C*[u^(C-1)]*u'
and
[C^x]'=[C^x]*log(C)
or using the chain rule
[C^v]'=[C^v]*log(C)*v'
inserting these into
[u^v]'=[u^C]'|C=v+[C^v]'|C=u
yeilds
[u^v]'=C*[u^(C-1)]*u'|C=v+[C^v]'+[C^v]*log(C)*v'|C=u
[u^v]'=v*[u^(v-1)]*u'+[u^v]*log(u)*v'