# How to find first derivative of function y=x \ln(x) by

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

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Charles Benedict
$$\displaystyle\lim_{{{h}\to\infty}}{\frac{{{\left({x}+{h}\right)}{\ln{{\left({x}+{h}\right)}}}-{x}{\ln{{x}}}}}{{{h}}}}=\lim_{{{h}\to\infty}}{\frac{{{x}{\ln{{\left({\frac{{{x}+{h}}}{{{x}}}}\right)}}}+{h}{\ln{{\left({x}+{h}\right)}}}}}{{{h}}}}$$
$$\displaystyle={x}{\ln{{\left[\lim_{{{h}\to\infty}}{\left({1}+{\frac{{x}}{{h}}}\right)}^{{{\frac{{1}}{{h}}}}}\right]}}}+\lim_{{{h}\to\infty}}{\ln{{\left({x}+{h}\right)}}}={1}+{\ln{{x}}}$$
where $$\displaystyle{x}{\ln{{\left[\lim_{{{h}\to\infty}}{\left({1}+{\frac{{x}}{{h}}}\right)}^{{{\frac{{1}}{{h}}}}}\right]}}}={1}$$ follows from the well known limit:
$$\displaystyle\lim_{{{h}\to\infty}}{\left({1}+{\frac{{x}}{{h}}}\right)}^{{{\frac{{1}}{{h}}}}}={e}^{{{\frac{{1}}{{x}}}}}$$
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Karen Robbins
We have that
$$\displaystyle\lim_{{{h}\to{0}}}{\frac{{{\left({x}+{h}\right)}{\log{{\left({x}+{h}\right)}}}-{x}{\log{{x}}}}}{{{h}}}}=\lim_{{{h}\to{0}}}{\frac{{{x}{\left({\log{{\left({x}+{h}\right)}}}-{\log{{x}}}\right)}+{h}{\log{{\left({x}+{h}\right)}}}}}{{{h}}}}=$$
$$\displaystyle=\lim_{{{h}\to{0}}}{\left({x}\cdot{\frac{{{\log{{\left({x}+{h}\right)}}}-{\log{{x}}}}}{{{h}}}}+{\log{{\left({x}+{h}\right)}}}\right)}={x}\cdot{\frac{{1}}{{x}}}+{\log{{x}}}$$
indeed
$$\displaystyle{\frac{{{\log{{\left({x}+{h}\right)}}}-{\log{{x}}}}}{{{h}}}}={\frac{{1}}{{x}}}{\frac{{{\log{{\left({1}+{\frac{{h}}{{x}}}\right)}}}}}{{{\frac{{{h}}}{{{x}}}}}}}\to{\frac{{1}}{{x}}}$$
indeed by $$\displaystyle{y}={\frac{{x}}{{h}}}\to\infty$$
$$\displaystyle{\frac{{{\log{{\left({1}+{\frac{{h}}{{x}}}\right)}}}}}{{{\frac{{h}}{{x}}}}}}={{\log{{\left({1}+{\frac{{1}}{{y}}}\right)}}}^{{y}}\to}{\log{{e}}}={1}$$
Vasquez

Continuing with the suggestion in my comment, you might do
$$\begin{array}{}\lim_{h \to 0} \frac{(x+h)\ln(x+h)-x \ln x}{h}=\lim_{h \to 0} \frac{(x+h)\ln(x+h)-x \ln(x+h)+x \ln(x+h)-x \ln x}{h} \\=(\lim_{h \to 0} \ln(x+h))(\lim_{h \to 0} \frac{(x+h)-x}{h})+(\lim_{h \to 0} x)(\lim_{h \to 0} \frac{\ln(x+h)-\ln x}{h}) \\=\ln x \lim_{h \to 0} \frac{(x+h)-x}{h}+x \lim_{h \to 0} \frac{\ln(x+h)-\ln x}{h} \end{array}$$
x+h−x=h , so the first limit is 1.

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