If y= f(x)= x^x, evaluate f'(e)

[Guest]

If y= f(x)= x^x, evaluate f'(e)

how do you do this?
thanks

 

[ChaoticLlama]

First step is to 'ln' both sides so that you can bring down the exponent.

\(\displaystyle \L\begin{array}{c}
y = x^x \\
\ln y = \ln x^x \\
\ln y = x\ln x \\
\end{array}\)

can you finish from here? (all you have to do is differentiate)

 

[galactus]

Use logarithmic differentiation.

\(\displaystyle \L\\y=x^{x}\)

Take log of both sides:

\(\displaystyle \L\\ln(y)=xln(x)\)

Differentiate:

\(\displaystyle \L\\\frac{dy}{y}=ln(x+1)dx\)

But, \(\displaystyle \L\\y=x^{x}\)

\(\displaystyle \L\\\frac{dy}{dx}=ln(x+1)x^{x}\)

Now, sub in e.

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
y & = & x^x \\
\ln (y) & = & x\ln (x) \\
\frac{{y'}}{y} & = & \ln (x) + x(1/x)\quad \mbox{Note the product!} \\
y' & = & y\left[ {\ln (x) + 1} \right] \\
y' & = & \left( {x^x } \right)\left[ {\ln (x) + 1} \right] \\
\end{array}\)

 

[stapel]

If y= f(x)= x^x, evaluate f'(e)

how do you do this?

Just like I said in your previous thread, this is where you use logarithmic differentiation.

Eliz.