What is the derivative of X^(X) and please explain how this is done.?

Question

What is the derivative of X^(X) and please explain how this is done.

Many Thanks.

Answer 1

logarithmic differentiation

y=x^x
log each side
ln y = ln(x^x)
simplify
lny = x lnx
derivative with product rule
1/y * y' = (1) ln x + x (1/x)
1/y * y ' = ln x + 1
y ' = y * (ln x +1 ) sub in y=
y ' = x^x (ln x +1)

that is it!

:)

Answer 2

y=x^x

You cannot use the power rule or any of the other derivative rules since there is no rule for taking the derivative of a variable raised to a variable.

First, you have to rewrite the problem by taking the natural logs of both sides.

lny=ln(x^x)

Using the laws of logarithms, you can rewrite the problem as

lny=xlnx

Now, you can take the derivative of this using implicit differentiation on the left side and the product rule on the right side:

(1/y)(dy/dx) = x(1/X) + lnx(1) = 1+lnx

To solve for dy/dx, divide each side by (1/y) [or just mulitiply by y...much easier to think about.]:

dy/dx = y + ylnx

Since it looks funny to have your derivative by a function of both x and y, substitue back in for y from your original equation.

dy/dx = (x^(x) + x^(x) lnx

Since it is so easy to do so, just factor your answer.

dy/dx = x^(x)(1+lnx)

And that's the answer. :D

Answer 3

Let y = x^x.
We are going to use logarithmic differentiation
to find dy/dx.
Take the log of both sides
log y = x log x
1/y * dy/dx = 1 + log x.
Here we used the chain rule on the left-hand side
and the product rule on the right.
So dy/ dx = y(1 + log x) = x^x(1 + log x).

Answer 4

y=x^x and we want to find dy/dx

Take logarithm of both sides

ln y = x ln x

Take derivative of both sides with respect to x.

1/y dy/dx = ln x + 1

I used the chain rule on the left and the product rule on the right.

Now multiply both sides by y.

dy/dx= y(1 + ln x)

Now substitute y=x^x

dy/dx= x^x (1 + ln x)

Answer 5

Let's f(x) = x^(x)

ln f(x) = ln x^(x) (*ln x = logarithm with base e of x)
ln f(x) = x ln x (*formula log a^n = n log a )

derive both sides, get
1/f(x) * f'(x) = ln x (1) + x (1/x)
(*formula if g(x) = log x <=> g'(x) = 1/x )
(*formula if g(x) = u(x)v(x) <=> g'(x) = v(x) * u '(x) + u(x) * v '(x) )
1/f(x) * f'(x) = ln x + x
f'(x) = f(x) (ln x + x)
f'(x) = (x^x)(ln x + x) (* f(x) = x^x )
So the answer is (x^x)(ln x + x) #End#

Answer 6

Take ln
ln y = x*ln x
y´/y= lnx+1
so y´= x^x(ln x+1)

Answer 7

http://www.analyzemath.com/calculus/Differentiation/first_derivative.html
perfectly explained right there