If f(x) = (1 + 1/x)^x for x > 0, what is f'(1)?

Question 104

HardOriginal question

If $f(x) = \left(1 + \frac{1}{x}\right)^{x}$ for $x > 0$ , what is $f'\left(1\right)$ ?

-1

Correct answerYour answer

\log 2 - \frac{1}{2}

Correct answerYour answer

2\log 2 - 1

Correct answerYour answer

2\log 2

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$\log y = x\left[\log(x + 1) - \log x\right]$ , so $\frac{y'}{y} = \log\left(1 + \frac{1}{x}\right) + x\left(\frac{1}{x + 1} - \frac{1}{x}\right) = \log\left(1 + \frac{1}{x}\right) - \frac{1}{x + 1}$ .
At 1: $y = 2$ and $\frac{y'}{y} = \log 2 - \frac{1}{2}$ , so $f'(1) = 2\log 2 - 1$ .

Worth noting

$-1$ is the power rule $x\left(1 + \frac{1}{x}\right)^{x - 1} \cdot \left(-\frac{1}{x^{2}}\right)$ at 1, which treats the power $x$ as a constant.

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