Derivative of x to the x

[lapenna]

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

 

[theperthvan]

Without the derivative definition, ln both sides, so:

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

 

[HallsofIvy]

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!

 

[dextercioby]

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:

$$x^x =e^{x\ln x}$$

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

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

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

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

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

$$=x^x \left(\ln x +1\right)$$

 

[theperthvan]

wow .

 

[VietDao29]

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

$$x^x =e^{x\ln x}$$

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

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

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

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

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

$$=x^x \left(\ln x +1\right)$$

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

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

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

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

$$= x ^ x (1 + \ln (x))$$

 

[Raiey]

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

 

[g_edgar]

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

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

 

[Raiey]

An answer is attached in PDF