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

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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.
  • #1
lapenna
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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
 
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  • #2
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]
 
  • #3
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!
 
  • #4
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:

[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]
 
Last edited:
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Likes Cristopher and Cristopher1
  • #5
wow .
 
  • #6
dextercioby said:
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]

Whoa. :bugeye:
Or somewhat easier. That's a little bit messy methinks: o:)
[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: (eu)'u = eu]

[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]
 
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  • #7
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
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  • #8
Raiey said:
Since the original equation is x^x and never a division quotient, one can not apply that rule.

"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.
 
  • #9
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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