A friend and I were messing around with derivatives, and while we both know the procedure for finding the derivative of $y=x^x$ with logarithmic differentiation, i.e.

$$y=x^x\\ ln(y)=x*ln(x)\\ \dfrac{dy}{dx}\dfrac{1}{y}=1+ln(x) \\ \dfrac{dy}{dx}=y(1+ln(x))\\ \dfrac{dy}{dx}=x^x(1+ln(x))$$

when we tried to do it with the formal definition of the derivative, we got this

$$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\\ \lim_{h \to 0}\frac{x^{x+h}-x^x}{h}\\ \lim_{h \to 0}\frac{x^xx^h-x^x}{h}\\ \lim_{h \to 0}\frac{x^x(x^h-1)}{h}$$ then we pulled out the $x^x$ $$x^x\lim_{h \to 0}\frac{x^h-1}{h}$$

Then, using l'Hopital's rule, we differentiated with respect to $h$, like this

$$\lim_{h \to 0} \frac{x^h-1}{h}\\ \lim_{h \to 0} \frac{x^h*ln(x)}{1}$$

and since $x^h$ approaches zero, we get $ln(x)$ for the limit

Putting that together, we have

$$\dfrac{dy}{dx}=x^x ln(x)$$

and somewhere in between, we lost an $x^x$. Where did I go wrong here? Am I not allowed to take the derivative with only respect to $h$?

  • 10
    $\begingroup$ $f(x+h)=(x+h)^{x+h}$, not $x^{x+h}$. $\endgroup$ – blue Mar 2 '14 at 10:31
  • $\begingroup$ @seaturtles ohhhh. That would make sense $\endgroup$ – scrblnrd3 Mar 2 '14 at 10:33

Here are the steps $$ \frac{d}{dx}x^x=\lim_{h\to 0} \frac{(x+h)^{x+h}-x^x}{h} =\lim_{h\to 0} \frac{\frac{d}{dh}(x+h)^{x+h}-\frac{d}{dh}x^x}{\frac{d}{dh}h} =\lim_{h\to 0} \frac{\frac{d}{dh}e^{\ln(x+h)^{x+h}}-0}{1} = \lim_{h\to 0} \frac{d}{dh}e^{(x+h)\ln(x+h)}=\lim_{h\to 0} e^{(x+h)\ln(x+h)}\frac{d}{dh}[(x+h)\ln(x+h)]= \lim_{h\to 0} (x+h)^{x+h}\left((x+h)\frac{d}{dh}\ln(x+h)+\ln(x+h)\frac{d}{dh}(x+h)\right) = \lim_{h\to 0} (x+h)^{x+h}\left(\frac{x+h}{x+h}+\ln(x+h)\right) = \lim_{h\to 0} (x+h)^{x+h}\left(1+\ln(x+h)\right)= x^x(1+\ln x)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.