Question

How do you differentiate `y=log_x(2)`?

Answer 1

`x^y = 2`

`ln(x^y) = ln2`

`yln(x) = ln2`

By the product rule and the differential identity `(ln(x))' = 1/x`:

`1(dy/dx)lnx + y(1/x) = 0`

`lnx(dy/dx) = -y/x`

`dy/dx = (-y/x)/lnx`

`dy/dx = -y/(xlnx)`

Since `y = ln_x(2)`:

`dy/dx =- (ln_x(2))/(xlnx)`

We can simplify this further using the rule `log_a(n) = logn/loga`.

`dy/dx = -(ln2/lnx)/(xlnx)`

`dy/dx =- ln2/(x(lnx)^2)`

Hopefully this helps!