Derivative of x to the x

Without the derivative definition, ln both sides, so:

[tex]y=x^x[/tex]
[tex]ln(y)=xln(x)[/tex]
[tex]\frac{1}{y}.\frac{dy}{dx} = 1.ln(x) + x.\frac{1}{x}[/tex]
[tex]\frac{dy}{dx} = y[ln(x) + 1][/tex]
[tex]\frac{dy}{dx} = x^x(ln(x) + 1) [/tex]

 

I can't imagine there being any reasonable way to use the "difference quotient" to determine the derivative of x^x. You must have a really evil teacher!

 

Either this is unreasonable, or your imagination really needs to see more:

[tex] x^x =e^{x\ln x} [/tex]

[tex] \lim_{h\rightarrow 0} \frac{(x+h)^{(x+h)}-x^x}{h} =\lim_{h\rightarrow 0} \frac{e^{(x+h)\ln(x+h)}-e^{x\ln x}}{h}=\lim_{h\rightarrow 0}\frac{e^{(x+h)\ln\left(x\left(1+\frac{h}{x}\right)\right)}-e^{x\ln x}}{h}[/tex]

[tex] =\lim_{h\rightarrow 0} \frac{e^{x\ln x}e^{h\ln x}e^{x\ln\left(1+\frac{h}{x}\right)}e^{h\ln\left(1+\frac{h}{x}\right)}-e^{x\ln x}}{h} [/tex]

[tex] =x^{x}\lim_{h\rightarrow 0}\frac{e^{h\ln x}e^{x\ln\left(1+\frac{h}{x}\right)}e^{h\ln\left(1+\frac{h}{x}\right)}-1}{h}\cdot \frac{h\ln x+x\ln\left(1+\frac{h}{x}\right)+h\ln\left(1+\frac{h}{x}\right)}{h\ln x+x\ln\left(1+\frac{h}{x}\right)+h\ln\left(1+\frac{h}{x}\right)} [/tex]

[tex] =x^{x}\left[\ln x+x\lim_{h\rightarrow 0}\frac{\ln\left(1+\frac{h}{x}\right)}{h}+\lim_{h\rightarrow 0}\ln\left(1+\frac{h}{x}\right)\right] [/tex]

[tex] =x^x \left\{\ln x+x\ln \lim_{h\rightarrow 0}\left[\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}\right\} [/tex]

[tex] =x^x \left(\ln x +1\right) [/tex]

 

wow .

 

Whoa.
Or somewhat easier. That's a little bit messy methinks:
[tex](x ^ x)' = (e ^ {x \ln (x)})' = \lim_{h \rightarrow 0} \frac{e ^ {(x + h) \ln (x + h)} - e ^ {x \ln (x)}}{h}[/tex]

[tex]= \lim_{h \rightarrow 0} \left( \frac{e ^ {(x + h) \ln (x + h)} - e ^ {x \ln (x)}}{(x + h) \ln (x + h) - x \ln x} \times \frac{(x + h) \ln (x + h) - x \ln x}{h} \right)[/tex]

[tex]= e ^ {x \ln (x)} \left( \lim_{h \rightarrow 0} \frac{x \ln (1 + \frac{h}{x}) + h \ln (x + h)}{h} \right)[/tex] [due to: (e^u)'_u = e^u]

[tex]= x ^ x \left\{ \lim_{h \rightarrow 0} \left[ x \ln \left( \left( 1 + \frac{h}{x} \right) ^ {\frac{x}{h}} \right) ^ {\frac{1}{x}} \right] + \ln (x) \right\}[/tex]

[tex]= x ^ x (1 + \ln (x))[/tex]

 

Since the original equation is x^x and never a division quotient, one can not apply that rule. Here is the alternative proof.

 

"Difference quotient" means this:
[tex]
\frac{f(x)-f(a)}{x-a}
[/tex]
so the assignment was to use the definition of derivative that involves this.

 

An answer is attached in PDF