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Homework Statement
I am required to find the derivative of x^x using the differance quotient
Homework Equations
((x+h)^(x+h)-(x^x))/h
The Attempt at a Solution
TI-89 gives (lnx+1)x^x I can't figure out how to get there
I can't imagine there being any reasonable way to use the "difference quotient" to determine the derivative of xx. 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]
Since the original equation is x^x and never a division quotient, one can not apply that rule.