Question

Can somebody explain the mechanic of why the derivative of x^x = x^x(ln(x) + 1) ...thanks

Answer 1

let's just do the right side

x^x----> x ln(x)

1+ ln(x)

Answer 2

you would want to use logarithmic differentiation on:\[y=x^x\]like we did in that last problem you asked.

Answer 3

the derivative of x^x is really neat, it is like a hybrid of the exponential rule:
c^x --> c^x(lnc)
and the power rule:
x^n --> nx^(n-1)
because if you distribute the answer above: x^x(lnx + 1), you get
x^x * lnx + x^x
If you were to simply treat x^x like an exponential function, you would get the first term
x^x * lnx
and if you were to treat x^x like a power function, you get the second term because n is x:
x^x --> x*x^(x-1) = x^x
It may be misleading to point this out, but I find it reassuring.

Answer 4

Of course the way you get it isn't by adding the two at all, it is by logarithmic differentiation, start with y = x^x and take the natural log of both sides:
lny = lnx^x
use rules of logs:
lny = xlnx
take derivative, using the chain rule on the left and the product rule on the right:
y'/y = x*1/x + lnx
simplify:
y'/y = 1 + lnx
multiply through by y, which is x^x:
y' = x^x(1+lnx)

Answer 5

the meaning of
\[x^x\] is
\[e^{x\ln(x)}\] now take the derivative using the chain rule. all the work is in the exponent, since the derivative of
\[e^{\text{ something }}\] is
\[e^ {\text{ something }}\times \frac{d}{dx}\text{ something}\]