Single Variable Calculus

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Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 32

Expert Answers

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Differentiate $\displaystyle f(x) = \frac{1}{1 + \ln x}$ and find the domain of $f$

The denominator of the given function should be greater than zero. So,

$ \begin{equation} \begin{aligned} 1 + \ln x & > 0 \\ \\ \ln x & > - 1\\ \\ e^{\ln x} & > e^{-1}\\ \\ x &> e^{-1}\\ \\ x &> \frac{1}{e} \end{aligned} \end{equation} $

Therefore, the domain is $\displaystyle \left[ 0, \frac{1}{e} \right)\bigcup \left( \frac{1}{e}, \infty \right)$

Solving for $f'$

$ \begin{equation} \begin{aligned} f'(x) &= \frac{d}{dx} \left( \frac{1}{1 + \ln x} \right)\\ \\ f'(x) &= \frac{(1+ \ln x) \frac{d}{dx} (1) - (1) \frac{d}{dx}(1 + \ln x)}{(1+ \ln x)^2}\\ \\ f'(x) &= \frac{-\frac{1}{x}}{(1+\ln x)^2}\\ \\ f'(x) &= \frac{-1}{x(1 + \ln x)^2} \end{aligned} \end{equation} $

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