Can one differentiate xx (x to the power of
x)
Thank you
Sarah Shales
Hi:
You can. You'll need the chain rule (for differentiating a
function of a function)
Sometimes things like ab look much easier if you take
logs:
y = xx
=> log y = x log(x)
So, using the chain rule on the LHS:
(1/y) (dy/dx) = log(x) + x(1/x)
Thus
dy/dx = xx (log(x) + 1)
Hope this helps,
Stuart
I'm not too great with the binomial theorem, so the
differential of
y=xx
has been puzzling me all day.
What is the differential, and how is it found?
The way to do it is to write it in a way
that you know the answer.
You know that the differential of ef(x) is
f'(x)ef(x) (that's the chain rule).
Also, xx =(eln(x) )x
=ex.ln(x) , so now you need to work out the
differential of x.ln(x).
Write back again if you still can't get it.
I know that the differential of x.lnx is 1+lnx, but Im having trouble proving this.
Do you know the product rule, that if
f(x)=u(x)v(x), then f'(x)=u'(x)v(x)+u(x)v'(x). So, if you put u=x
and v=ln(x), you should get the answer. Also, as you probably
know, the differential of ln(x) is 1/x.
Does that answer your question? I can prove the results above if
you want (e.g. chain rule, product rule, (lnx)'=1/x, etc.)
I know how to prove all the results you mentioned spare the product rule. How does the proof go?
Well, this is an almost proof;
f+df=(u+du)(v+dv)=uv+v.du+u.dv+du.dv=f+v.du+u.dv+du.dv (as
f=uv).
Subtract f from both sides to get df=u.dv+v.du+du.dv, divide by
dx to get df/dx=u.dv/dx+v.du/dx+du.dv/dx, the last term is
du.dv/dx which is 0, sort of.
Therefore df/dx=u.dv/dx+v.du/dx, or f'=uv'+vu'.
Tada!
[More on this here .
- The Editor]
Alternatively, use logarithms:
y = uv
where y,u,v are functions of x.
ln y = ln(u) + ln(v)
Differentiate:
1/y dy/dx = 1/u du/dx + 1/v dv/dx
(Using the chain rule and assuming that ln(x) differentiates to
1/x. Both are easy to prove.)
dy/dx = y/u du/dx + y/v dv/dx
dy/dx = v du/dx + u dv/dx. So:
d(uv)/dx = v du/dx + u dv/dx.
That's a better proof, thanks Michael.