# How do I find the derivative of x^x using the difference quotient?

• lapenna
In summary, the conversation involves finding the derivative of x^x using the difference quotient and the attempt at a solution involves using the definition of derivative and performing various mathematical steps to arrive at the answer of x^x(1+ln(x)). The conversation also includes a side note about an alternative proof for finding the derivative of x^x.
lapenna

## Homework Statement

I am required to find the derivative of x^x using the difference 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

Without the derivative definition, ln both sides, so:

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

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!

HallsofIvy said:
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:

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

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

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

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

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

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

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

Last edited:
Cristopher and Cristopher1
wow .

dextercioby said:
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: (eu)'u = eu]

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

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

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

#### Attachments

• derivativesxtopowerofx.txt
3.4 KB · Views: 384
Raiey said:
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.

An answer is attached in PDF

#### Attachments

• dydxofxraisedtox.pdf
37.7 KB · Views: 232

## 1. What is the derivative of x to the x?

The derivative of x to the x is given by the formula xx-1(1+ln(x)).

## 2. How do you find the derivative of x to the x?

To find the derivative of x to the x, you can use the power rule and logarithmic differentiation. First, rewrite x to the x as exln(x), then use the power rule to find the derivative of exln(x). Finally, use logarithmic differentiation to simplify the expression.

## 3. What is the significance of the derivative of x to the x?

The derivative of x to the x is used in many areas of mathematics, including calculus, differential equations, and complex analysis. It is also used in applications such as physics and economics to model exponential growth and decay.

## 4. Can the derivative of x to the x be negative?

Yes, the derivative of x to the x can be negative. This occurs when x is between 0 and 1, as the negative value of the logarithm term outweighs the positive value of the power term in the derivative formula.

## 5. How does the derivative of x to the x compare to other exponential functions?

The derivative of x to the x is unique compared to other exponential functions. While the derivative of ax is axln(a), the derivative of x to the x involves both a power and a logarithm term, making it more complex to calculate.

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