3
$\begingroup$

How to find first derivative of function $y=x \ln(x)$ by limit definition, that is using this formula $$y'=\lim_{h\to 0}\frac{f(x+h)-f(x)}h$$

not product rule or L'Hopital rule are allowed.

Thank you in advance

$\endgroup$
2
  • $\begingroup$ Without directly using the product rule, you can start by using the same "trick" often used in the derivative of the rule:$$\lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln x}h=\lim_{h\to0}\frac{(x+h)\ln(x+h)\color{red}{{}-x\ln(x+h)+x\ln(x+h)}-x\ln x}h$$ $\endgroup$ – user170231 Oct 14 '20 at 14:20
  • $\begingroup$ @user170231 thank you so much for your response but how can i find from your last step that first derivative of y=x*ln(x) is y'=ln(x)+1 ? :) $\endgroup$ – John Oct 14 '20 at 14:30
2
$\begingroup$

We have that

$$\lim_{h \to 0}\frac{(x+h)\log (x+h)-x\log x}{h}=\lim_{h \to 0}\frac{x\left(\log (x+h)-\log x\right)+h\log (x+h)}{h}=$$

$$=\lim_{h \to 0} \left(x \cdot \frac{\log (x+h)-\log x}{h} +\log (x+h)\right)=x \cdot \frac1x +\log x$$

indeed

$$\frac{\log (x+h)-\log x}{h}=\frac1x\frac{\log \left(1+\frac hx\right)}{\frac hx} \to \frac1x$$

indeed by $y= \frac x h \to \infty$

$$\frac{\log \left(1+\frac hx\right)}{\frac hx}=\log \left(1+\frac1y\right)^y \to \log e=1$$

$\endgroup$
4
  • $\begingroup$ thank you so much for your response but how can i find from your last step that first derivative of y=x*ln(x) is y'=ln(x)+1 ? :) , i mean how to go to the result y'=ln(x)+1 :) $\endgroup$ – John Oct 14 '20 at 14:31
  • $\begingroup$ @John Are you given that $ \lim_{h \to 0} \frac{\log (x+h)-\log x}{h}=\frac1x$? $\endgroup$ – user Oct 14 '20 at 14:33
  • $\begingroup$ Thank you so much sir now i completely understand problem and solution.You were so fast and accurate to this problem :) $\endgroup$ – John Oct 14 '20 at 14:52
  • $\begingroup$ @John You are very welcome! Bye $\endgroup$ – user Oct 14 '20 at 15:03
3
$\begingroup$

\begin{multline} \lim_{h\to\infty}\frac{(x+h)\ln(x+h)-x\ln x}{h}=\lim_{h\to\infty}\frac{x\ln\left(\frac{x+h}{x}\right)+h\ln(x+h)}{h}\\=x\ln\left[\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h\right]+\lim_{h\to\infty}\ln(x+h)=1+\ln x \end{multline} where $x\ln\left[\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h\right]=1$ follows from the well known limit: $$\lim_{h\to\infty}\left(1+\frac{x}{h}\right)^\frac1h=e^{\frac1x}$$

$\endgroup$
1
  • $\begingroup$ Thank you so much for your answer :) $\endgroup$ – John Oct 14 '20 at 15:21
2
$\begingroup$

Continuing with the suggestion in my comment, you might do

$$\begin{align} \lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln x}h&=\lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln(x+h)+x\ln(x+h)-x\ln x}h\\[1ex] &=\left(\lim_{h\to0}\ln(x+h)\right)\left(\lim_{h\to0}\frac{(x+h)-x}h\right)+\left(\lim_{h\to0}x\right)\left(\lim_{h\to0}\frac{\ln(x+h)-\ln x}h\right)\\[1ex] &=\ln x\lim_{h\to0}\frac{(x+h)-x}h+x\lim_{h\to0}\frac{\ln(x+h)-\ln x}h \end{align}$$

$(x+h)-x=h$, so the first limit is $1$. For the other limit, see the other answers here, or methods shown here.

$\endgroup$
1
  • $\begingroup$ Thank you so much for your answer :) $\endgroup$ – John Oct 14 '20 at 14:55

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.