Derivative of 1/lnx

[ZeroOne23]

Homework Statement

Find the derivative of 1 / ln x

Homework Equations

N/A

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? =/

 

[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)

 

[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.

 

[JasonRox]

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

 

[z-component]

What made you doubt your first attempt? Using the product rule:

[tex]\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 }}[/tex]

 

[HallsofIvy]

That isn't the product rule!

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

 

[Smartass]

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