- #1
lapenna
- 1
- 0
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
HallsofIvy said: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 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]
Raiey said:Since the original equation is x^x and never a division quotient, one can not apply that rule.
The derivative of x to the x is given by the formula x^{x-1}(1+ln(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 e^{xln(x)}, then use the power rule to find the derivative of e^{xln(x)}. Finally, use logarithmic differentiation to simplify the expression.
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.
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.
The derivative of x to the x is unique compared to other exponential functions. While the derivative of a^{x} is a^{x}ln(a), the derivative of x to the x involves both a power and a logarithm term, making it more complex to calculate.