Question

What is the derivative of `1/logx`?

Answer 1

It is `(-1)/(x(logx)^2(ln10)` (Assuming that `logx = log_10x`).

Explanation:

The function `f(x) = 1/logx`

is of the form: `f(x) = 1/u = u^-1`

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

`f'(x) = -1u^-2 * (du)/dx = (-1)/u^2 * (du)/dx`

In this case, `u = logx`, which I take to mean `log_10x`.

The derivative of `log_bx = 1/(xlnb)`, so we have:

`(du)/dx d/dx(logx) = d/dx(log_10 x)= 1/(xln10)`.

Putting this together, we get:

`f'(x) = (-1)/(logx)^2 * 1/(xln10) = (-1)/(x(logx)^2(ln10)`

Note
If we are using `logx` 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) = (-1)/(logx)^2 * 1/x = (-1)/(x(logx)^2)`