# What is the minimum value of the function y=e^ln(x^x) ?

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### 2 Answers

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))

=> (1/e)^(1/e)

=> 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

f(1/e)=e^[ln(1/e)^1/e]

f(1/e)=e^(-1/e)

**The minimum point is represented by it's coordinates (1/e, e^(-1/e)).**