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

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.

• Without directly using the product rule, you can start by using the same "trick" often used in the derivative of the rule:$$\lim_{h\to0}\frac{(x+h)\ln(x+h)-x\ln x}h=\lim_{h\to0}\frac{(x+h)\ln(x+h)\color{red}{{}-x\ln(x+h)+x\ln(x+h)}-x\ln x}h$$ – user170231 Oct 14 '20 at 14:20
• @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 ? :) – John Oct 14 '20 at 14:30

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

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

• 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 :) – John Oct 14 '20 at 14:31
• @John Are you given that $\lim_{h \to 0} \frac{\log (x+h)-\log x}{h}=\frac1x$? – user Oct 14 '20 at 14:33
• Thank you so much sir now i completely understand problem and solution.You were so fast and accurate to this problem :) – John Oct 14 '20 at 14:52
• @John You are very welcome! Bye – user Oct 14 '20 at 15:03

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

• Thank you so much for your answer :) – John Oct 14 '20 at 15:21

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.

• Thank you so much for your answer :) – John Oct 14 '20 at 14:55