Derivative of 1 / ln x

1. Oct 28, 2007

ZeroOne23

1. The problem statement, all variables and given/known data

Find the derivative of 1 / ln x

2. Relevant equations

N/A

3. The attempt at a solution

y = 1/lnx

First Attempt:
y' = -1/x/(lnx)^2
y' = -1 / x(lnx)^2

Second Attempt:
ln y = ln (1 / lnx)
ln y = ln 1 - ln x
ln y = -lnx
dy/dx = y(-1/x)
dy/dx = -1/xlnx

Third Attempt:
ln y = -lnx
y = -x
y' = -1

Which one is it? =/

2. Oct 28, 2007

rock.freak667

In your second attempt this step is wrong
ln y = ln (1 / lnx)
ln y = ln 1 - ln x

It should be lny=ln 1 -ln(lnx)

3. Oct 28, 2007

hotcommodity

We know the derivative would be the denominator times the derivative of the numerator(which would be zero in this case), minus the numerator times the derivative of the denominator(which is 1/x), over the denominator squared.

4. Oct 28, 2007

JasonRox

Use the product rule. Sure you can do the quotient rule but the product rule is so easy to remember!

5. Oct 28, 2007

z-component

$$\frac{d}{{dx}}\left( {\frac{1}{{\ln x}}} \right) = \frac{{0 \cdot \ln x - 1\left( {\frac{1}{x}} \right)}}{{\left( {\ln x} \right)^2 }} = \frac{{ - \frac{1}{x}}}{{\left( {\ln x} \right)^2 }} = - \frac{1}{{x\left( {\ln x} \right)^2 }}$$

6. Oct 29, 2007

HallsofIvy

Staff Emeritus
That isn't the product rule!

What JasonRox meant, I think, was use the chain rule on (ln x)-1.

7. Oct 29, 2007

Smartass

Your first approach was correct, the second one, as already pointed, instead of ln(lnx) you took lnx.