So $$\frac{dy}{dx}=(x)\left(\frac1x\right)+\ln(x)$$ $$=1+\ln(x).$$ This is one of the types of questions where the answer given may be wrong, but I would like to verify it anyway. The answer given is $$1-\ln(x).$$
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Have you typed the correct equation for $y$? – Shubham Johri Aug 13 '19 at 16:16
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4Did you mean $y = x ln x$? – Leo Aug 13 '19 at 16:16
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Question has been edited. It was a typo – Aditya Aug 13 '19 at 16:19
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Your answer is correct. – 19aksh Aug 13 '19 at 16:20
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Your book has a sign error: $$\frac{\partial (x \ln x)}{\partial x} = 1 + \ln x.$$ – David G. Stork Aug 13 '19 at 16:21
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@DavidG.Stork The OP has said that it is his book's answer – 19aksh Aug 13 '19 at 16:22
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1@Ak19: And the book is wrong. – David G. Stork Aug 13 '19 at 16:22
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1You should really change your book – Shubham Johri Aug 13 '19 at 16:22
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All these collective questions are being asked as I have noted down all the questions I have problems with and asked them in bulk. – Aditya Aug 13 '19 at 16:24
1 Answers
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Apply$$\frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$$
Where $f(x)=x, g(x)=\ln x$
It gives $$(x\ln x)'=1\times\ln x+x\times\frac{1}{x}=1+\ln x$$
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