Question

What is the derivative of `f(x) = 1/log_2x`?

Answer 1

`dy/dx = ln(2)/(x*ln^2(x))`

Explanation:

Using the change of base formula for logarithms we have

`log_2(x) = ln(x)/ln(2)`

So we have

`y = 1/(ln(x)/ln(2)) = ln(2)/ln(x)`

Differentiating we have

`dy/dx = ln(2)d/dx(1/ln(x))`

Calling `ln(x) = u`

`dy/dx = ln(2)d/(du)(1/u)d/dxln(x)`
`dy/dx = ln(2)*(-1/u^2)*1/x`

Substituting that back to terms of `x`

`dy/dx = ln(2)/(x*ln^2(x))`

Or, if you prefer

`dy/dx = 1/(x*log_2(x)*ln(x))`