ddx(1ln⁡(x)) Apply exponent rule =ddx(ln−1(x)) Now, apply The Chain Rule: =−1(ln⁡(x))2ddx(ln⁡(x)) =−1(ln⁡(x))2×1x Simplify −1xln2(x)

ddx(1ln⁡(x)) =ddx(ln−1(x)) Apply The Chain Rule: =−1(ln⁡(x))2ddx(ln⁡(x)) =−1(ln⁡(x))2×1x Now, we can simplify: −1xln2(x)

ddx(1ln⁡(x)) Apply exponent rule =ddx(ln−1(x)) Now, apply The Chain Rule: =−1(ln⁡(x))2ddx(ln⁡(x)) =−1(ln⁡(x))2×1x Simplify −1xln2(x)

ddx(1ln⁡(x)) =ddx(ln−1(x)) Apply The Chain Rule: =−1(ln⁡(x))2ddx(ln⁡(x)) =−1(ln⁡(x))2×1x Now, we can simplify: −1xln2(x)

"Integrate 1ln⁡(x)" was answered by our experts

Stefan Hendricks

Answered question

2021-12-11

Integrate

\frac{1}{\ln (x)}

Jeremy Merritt

Beginner2021-12-12Added 31 answers

\frac{d}{d x} (\frac{1}{\ln (x)})

Apply exponent rule

= \frac{d}{d x} (\ln^{- 1} (x))

Now, apply The Chain Rule:

= - \frac{1}{{(\ln (x))}^{2}} \frac{d}{d x} (\ln (x))

= - \frac{1}{{(\ln (x))}^{2}} \times \frac{1}{x}

Simplify

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

Esther Phillips

Beginner2021-12-13Added 34 answers

\frac{d}{d x} (\frac{1}{\ln (x)})

= \frac{d}{d x} (\ln^{- 1} (x))

Apply The Chain Rule:

= - \frac{1}{{(\ln (x))}^{2}} \frac{d}{d x} (\ln (x))

= - \frac{1}{{(\ln (x))}^{2}} \times \frac{1}{x}

Now, we can simplify:

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

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