Resolved: "What is the derivative of 1log⁡x?"

Kathleen Rausch

Answered question

2021-12-13

What is the derivative of

\frac{1}{\log x}

Virginia Palmer

Beginner2021-12-14Added 27 answers

It is

\frac{- 1}{x {(\log x)}^{2} (\ln 10)}

(Assuming that

\log x = \log_{10} x

Explanation:
The function

f (x) = \frac{1}{\log x}

is of the form:

f (x) = \frac{1}{u} = u^{- 1}

So we find its derivative using the power rule and the chain rule:

f ‘ (x) = - 1 u^{- 2} \cdot \frac{d u}{d x} = \frac{- 1}{u^{2}} \cdot \frac{d u}{d x}

In this case,

u = \log x which I take to mean \log_{10} x

The derivative of

\log_{b} x = \frac{1}{x \ln b}, so we have:

\frac{d u}{d x} \frac{d}{d x} (\log x) = \frac{d}{d x} (\log_{10} x) = \frac{1}{x \ln 10}

Putting this together, we get:

f^{'} (x) = \frac{- 1}{{(\log x)}^{2}} \cdot \frac{1}{x \ln 10} = \frac{- 1}{x {(\log x)}^{2} (\ln 10)}

Note: If we are using

\log x

for the natural logarithm (as is sometimes done), then, that would be

\log_{e} x

and natural log of e is 1 , so we get

f^{'} (x) = \frac{- 1}{{(\log x)}^{2}} \cdot \frac{1}{x} = \frac{- 1}{x {(\log x)}^{2}}

Natalie Yamamoto

Beginner2021-12-15Added 22 answers

Wow, cleverly you came up with it, I didn't even understand at the beginning, well done

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