What is the minimum value of the function y=e^ln(x^x) ?
We have the function y = e^ln(x^x)
take the log of both the sides
ln y = ln (e^ln(x^x))
=> ln y = ln(x^x)
=> ln y = x*ln x
differentiate both the sides
(1/y)dy/dx = ln x + x/x
dy/dx = y*(ln x + 1)
Equating dy/dx = 0
=> e^ln(x^x)*(ln x + 1) = 0
=> ln x + 1 = 0
=> ln x = -1
=> x = e^-1
For x = e^-1
f(x) = e^ln((1/e)^(1/e))
Therefore the required minimum value of the function is e^(-1/e)
We' ll differentiate the function with respect to x
f'(x) = (e^lnx^x)(xlnx)'
We'll apply product rule for (xlnx)':
(xlnx)' = x'*ln x + x*(lnx)'
(xlnx)' = ln x + 1
f'(x) = [e^(x*lnx)]*( lnx+1)
We'll put f'(x)=0 => [e^(x*lnx)]*(lnx+1) = 0
Since the factor e^lnx is positive => lnx+1=0
lnx+1=0 => lnx=-1 => x=e^-1=1/e
The critical point is x=1/e
To determine the minimum point, we'll substitute x by the value of critical point
The minimum point is represented by it's coordinates (1/e, e^(-1/e)).