Derivative of x^x?

Question

I need to find the derivative of x^x based on the definition of derivative. Using the limit as h->0. (f(x+h)-f(x))/h I know the derivative based solely on information but i need the step by step derivative based on the definition of derivative taking the limit when h goes to zero. Any help?

Answer 1

(x+h)^(x+h)=(x+h)^x*(x+h)^h
=x^x*(1+h/x)^x*(x+h)^h
=x^x*(1+h/x)^x*x^h*(1+h/x)^h
(1+h/x)^x=1+h+O(h^2)
(1+h/x)^h=1+O(h^2)
x^h=exp(h*ln(x))
=1+h*ln(x)+O(h^2)
(x+h)^(x+h)
=x^x*(1+h+O(h^2))
*(1+h*ln(x)+O(h^2))*(1+O(h^2))
=x^x*(1+h+h*ln(x)+O(h^2))
So
((x+h)^(x+h)-x^x)/h
=x^x*(h+h*ln(x)+O(h^2))/h
=x^x*(1+ln(x)+O(h))
Thus
lim{h->0}((x+h)^(x+h)-x^x)/h
=x^x*(1+ln(x))

Answer 2

I believe it would be VERY difficult to solve this (at least using the limit definition of a derivative) with a pencil and paper, so this is the first time on Answers that I have resorted to using a computer.

Here is how I solved it using Maple:

> f1:= x^x;
> f2:= (x+h)^(x+h);
> g := (f2-f1) / h;
> limit(g, h=0);

Here is the answer it spit out:
[(x^x) * ln(x)] + x^x


If you are simply trying to find the derivative, I recommend doing what one of the people above me did. If you need to use the limit definition to find it, use a computer!

Answer 3

It helps to look at a transformation of x^x.

x^x
= exp{ln(x^x)}
= exp{xln(x)}

d/dx x^x
= d/dx exp{xln(x)}
= exp{xln(x)} * d/dx {xln(x)} by the chain rule (d/dx f{g(x)} = f'{g(x)} * g'(x))
= exp{xln(x)} * [x * d/dx{ln(x)} + ln(x) * d/dx(x)] by the product rule. (d/dx f(x)g(x) = f(x)g'(x) + g(x)f'(x))
= exp{xln(x)} * {x * 1/x + ln(x) * 1}
= exp{xln(x)} * {1 + ln(x)}
= x^x * {1 + ln(x)}, x > 0

edit: Oh jeez, I didn't see that you needed to use the definition. I'll think about it.

edit: I don't care if you thumbs down me Dr. D, and I seriously don't care if you're right or wrong. I only care about this being fair to everyone. That's my whole problem with you. There are some serious irregularities with the voting on some of your questions. People don't just get three votes all of the sudden for answers that are either very terse or very wrong - not as often as it seems to happen with you.

And I have given you thumbs up in the past. I'm sure that you are very intelligent and are right more than 50% of the time, and I let my frustation with you spill over on that one question. So I apologize about that. In the end, all I care about is fairness. And I just have serious concerns about you. Let the voters make their choice.

Answer 4

It's tricky, so I am stealing this from another place....

y = x^x and x > 0 then ln y = ln (x^x)

ln y = x ln x

We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right.

y'(1/y) = ln x + x(1/x) = ln x + 1

Multiply both sides by y:

y ' = (ln x + 1)y

Finally, substitute x^x for y to obtain:

y' = (ln x + 1)x^x

Answer 5

Blahb31 had the correct answer.

You can check your answers using numerical calculus at Graphulator.com?=y=x^x

Answer 6

If you're allowed to take logs first,
y = x^x
z = ln(y) = x*ln(x)

Now z + dz = (x+dx) * ln(x+dx)
= (x+dx)* [ln(x) + ln(1 + dx/x)]
using the maclaurin series for ln(1 + dx/x),
z + dz = (x+dx)*[ln(x) + dx/x + O(dx^2)]
= x*lnx + dx + dx*ln(x) + O(dx^2)

dz = z+dz - z
= x*lnx - x*lnx + dx* (1 + lnx)
dz/dx 1/y * dy/dx = 1 + ln(x)
dy/dx = y*[1 + ln(x)]
= x^x * [1 + ln(x)]

**ALTERNATIVE APPROACH**
y = x^x
y + dy = (x+dx)^(x+dx)
Using the binomial expansion,
y + dy = x^(x+dx) + (x+dx)*x^(x+dx-1)*dx + O(dx^2)
= (x^x)*(x^dx) + dx*[x^(x+dx)] + O(dx^2)
= (x^x)*(x^dx) + [(x^x)*(x^dx)]*dx + O(dx^2)

At this point let's consider the behavior of a^h as h--> 0.
b = a^h
ln(b) = h*lna
b = e^(h*lna) ~= 1 + h*lna + O(h^2)
using maclauring expansion for e^(h*lna)
So x^dx ~= 1 + dx*lnx as dx --> 0

So y + dy = (x^x)*(1 + dx*lnx) + [(x^x)*(1 + dx*lnx)]*dx + O(dx^2)
dy = y+dy - y = RHS - x^x
= x^x * lnx * dx + x^x * dx + O(dx^2)
dy/dx = x^x * (lnx + 1)
Done