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

Adela Brown 2022-01-02 Answered
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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Expert Answer

Charles Benedict
Answered 2022-01-03 Author has 3288 answers
\(\displaystyle\lim_{{{h}\to\infty}}{\frac{{{\left({x}+{h}\right)}{\ln{{\left({x}+{h}\right)}}}-{x}{\ln{{x}}}}}{{{h}}}}=\lim_{{{h}\to\infty}}{\frac{{{x}{\ln{{\left({\frac{{{x}+{h}}}{{{x}}}}\right)}}}+{h}{\ln{{\left({x}+{h}\right)}}}}}{{{h}}}}\)
\(\displaystyle={x}{\ln{{\left[\lim_{{{h}\to\infty}}{\left({1}+{\frac{{x}}{{h}}}\right)}^{{{\frac{{1}}{{h}}}}}\right]}}}+\lim_{{{h}\to\infty}}{\ln{{\left({x}+{h}\right)}}}={1}+{\ln{{x}}}\)
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
Answered 2022-01-04 Author has 4657 answers
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 \(\displaystyle{y}={\frac{{x}}{{h}}}\to\infty\)
\(\displaystyle{\frac{{{\log{{\left({1}+{\frac{{h}}{{x}}}\right)}}}}}{{{\frac{{h}}{{x}}}}}}={{\log{{\left({1}+{\frac{{1}}{{y}}}\right)}}}^{{y}}\to}{\log{{e}}}={1}\)
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Vasquez
Answered 2022-01-09 Author has 9499 answers

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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