Question

### Gauthmathier9608

Grade 11 · 2021-01-10

The coefficient of (x-1)^{5} in the Taylor series for x \ln x about x=1 is （ ）

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Answer

4.7(568) votes

### Gauthmathier5583

Grade 11 · 2021-01-10

Answer

A

Explanation

f\left(x\right) = x \ln x, f'\left(x\right) = 1+ \ln x, f''\left(x\right) = \dfrac{1}{x}, f'''\left(x\right) = -\dfrac{1}{x^2}, f^{(4)}\left(x\right) = \dfrac{2}{x^3}, f^{(5)}\left(x\right)=-\dfrac {3\cdot 2}{x^{4}}, f^{(5)}\left(1\right)=-3\cdot 2

So the coefficient of (x-1)^{5} is -\dfrac {3\cdot 2}{5!}=-\dfrac {1}{20}.

So the coefficient of (x-1)^{5} is -\dfrac {3\cdot 2}{5!}=-\dfrac {1}{20}.

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