How to Differentiate y=x^x by First Principles?

[jiasyuen]

Anyone know how to differentiate \(\displaystyle y=x^x\) by first principles?

 

[Dethrone]

That requires using the definition of the derivative:

$$f'(a)=\lim_{{x}\to{a}}\frac{f(x)-f(a)}{x-a}$$

 

[jiasyuen]

I know this. But can you show me the solution? Thanks.

 

[Dethrone]

Are you trying to compute the derivative on a given point, or the general formula? I am saying this because the latter is quite tricky.

 

[jiasyuen]

Yea. Because I can't find the answer by using first principles.
The question also states that to simplify ur solution, assume x is integer and the function lie on first quadrant only (means no complex number here).

 

[Dethrone]

We can use either definitions:

$$f'(a)=\lim_{{x}\to{a}}\frac{x^x-a^a}{x-a}$$
or
$$f'(x)=\lim_{{h}\to{0}}\frac{(x+h)^{x+h}-x^x}{h}$$

The rest is just simplifying...what is the actual question, exactly? (Wondering)

 

[jiasyuen]

Differentiate \(\displaystyle y=x^x\) by using first principles. Assume x is integer and the function lie on first quadrant only.

Can you show the whole solution to me? Thanks.

 

[Dethrone]

$$f'(x)=\lim_{{h}\to{0}}\frac{(x+h)^{x+h}-x^x}{h}$$
$$=\lim_{{h}\to{0}}\frac{(x+h)^{x+h}-(x+h)^x+(x+h)^x-x^x}{h}$$
$$=\lim_{{h}\to{0}}\frac{(x+h)^x\left[(x+h)^h-1\right]}{h}+\lim_{{h}\to{0}}\frac{(x+h)^x-x^x}{h}$$
$$=x^x\ln\left({x}\right)+\lim_{{h}\to{0}}\frac{((x+h)-x)\left[x^{x-1}+x^{x-2}\cdot x+...+x\cdot x^{x-2}+x^{x-1}\right]}{h}$$

Can you finish this?

 

[jiasyuen]

The last step seems complicated to me. How to simplify it?
How the ln(x) comes from?

 

[Dethrone]

Notice that $(x+h)-x=h$, so how can you simply the fraction? Also, each term in the larger parenthesis simplifies to $x^{x-1}$, so we have $x$ terms of $x^{x-1}$.

It is known that $\lim_{{h}\to{0}}\frac{x^h-1}{h}=\ln\left({x}\right)$.

 

[jiasyuen]

So the final answer is \(\displaystyle x^x(\ln x+1)\)?

 

[Dethrone]

Yes, that is correct! (Cool)

 

[jiasyuen]

How to prove that \(\displaystyle \lim_{{h}\to{0}}\frac{x^h-1}{h}=\ln\left({x}\right)\)

 

[jiasyuen]

How to prove it by not using L'Hopital rule ?

 

[Dethrone]

Hi jiasyuen,

This is also not an easy limit to evaluate. Although it is tempting to use L'Hopital's rule or invoke the definition of the derivative, that is circular logic.

I am now unable to come up with a proof off the top of my head, and so if anyone has any ideas, feel free to comment.

 

[I like Serena]

Here's a different approach:
$$x^x = e^{\ln x^x} = e^{x\ln x}$$
Apply the chain rule followed by the product rule to get the derivative:
$$\d {} x \left({x^x}\right) = \d {} x \left(e^{x\ln x}\right) = e^{x\ln x}\cdot \d {} x(x\ln x) = e^{x\ln x}\cdot \left(1\cdot \ln x + x \cdot \frac 1 x\right) = x^x(\ln x + 1)$$

 

[jiasyuen]

Can anyone show it ? Thanks.

 

[Prometheus1]

Can anyone show it ? Thanks.

How is the natural logarithm defined in your course?