Question

What is the derivative of `f(x)=x/ln(1/x)`?

Answer 1

`d/dx x/(ln(1/x)) =(ln(1/x)+1)/(ln^2(1/x))`

Explanation:

To solve this, we will be using the quotient rule
`d/dx f(x)/g(x) = (f'(x)g(x) - f(x)g'(x))/g^2(x)`

and the chain rule
`d/dx f(g(x)) = f'(g(x))g'(x)`

For convenience, let's let `f_1(x) = x` and `f_2(x) = ln(1/x)`

First, we apply the quotient rule to get
`d/dx x/(ln(1/x)) = (f_1'(x)f_2(x) - f_1(x)f_2'(x))/f_2^2(x)`

We can easily see that `f_1'(x) = 1`, but we will need to do a little more work to calculate `f_2'(x)`

Before moving on, let's let `f_3(x) = ln(x)` and `f_4(x) = 1/x`

Then `f_2(x) = f_3(f_4(x))` and we can apply the chain rule to obtain
`f_2'(x) = f_3'(f_4(x))f_4'(x)`

`f_3'(x) = 1/x` and `f_4'(x) = -1/x^2`
so, substituting those in,

`f_2'(x) = (1/(1/x))(-1/x^2) = -1/x`

And now we can substitute back into the original equation to obtain

`d/dx x/(ln(1/x)) = (1*ln(1/x) - x*(-1/x))/(ln^2(1/x))=(ln(1/x)+1)/(ln^2(1/x))`

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While the above shows a general method, it is worth noting that in this case, we can save some time and effort by observing that
`ln(1/x) = ln(x^-1) = -ln(x)`

This removes the necessity of using the chain rule, as we can directly calculate the derivative
`d/dxln(1/x) = d/dx(-ln(x)) = -1/x`