`y` = `f(x)`
- ln
`y` = ln `f(x)`
`(d/dx)` (ln `y) = (d/dx) `(ln `f(x))`
- (1/
`y`) (`dy/dx`) = (`d/dx)` (ln `f(x)`)
`dy/dx` = `f(x) (d/dx)` (ln `f(x)`)

The first step is the given. In the second step, the natural logarithm is taken on both sides of the equation and then simplified using the Laws of Natural Logarithms. In the third step, both sides are implicitly differentiated. Step four is reached by theorem ((`d/dx`) (ln `u`) = (1/`u`) (`du/dx`) if `u` > 0). In step five, `f(x)` is multiplied by both sides, instead of `y` (`y` = `f(x)` ), so `y` is not part of the final derivative.

To complete the solution, ln `f(x)` must be differentiated at some stage after 3. It must also be pointed out that these guidelines are only for `f(x)` > 0 because guideline 2 would be undefined if `f(x)` < 0. If |`y`| = |`f(x)`| then the natural logarithms can be taken, obtaining ln |`y`| = ln |`f(x)`|. It can now be implicitly differentiated by the second part of the theorem described above ((`d/dx`) (ln |`u`|) = (1/`u`) ( `du/dx` if `u` != 0) can be used to arrive at guideline 4. Thus, negative values of `f(x)` do not change the outcome, and it is of no concern if `f(x)` is positive or negative. This method should not be used, however, to find `f'(a)` if `f(a)` = 0, since ln 0 is undefined.

Application for these guidelines would be to differentiate `f(x)` if it involves complicated products, quotients, or powers.

Example:

find `dy/dx` of
(5`x` - 4)^{3}
y = -----------------.
(2`x` + 1)^{(1/2)}
(guideline 2):
(5`x` - 4)^{3}
ln `y` = ln --------------------
(2`x` + 1)^{(1/2)}
ln `y` = ln((5`x` - 4)^{3}) - ln((2`x` + 1)^{(1/2)})
ln `y` = 3ln(5`x` - 4) - (1/2)ln(2`x` + 1)
(guideline 3):
(`d/dx`)ln `y` = (`d/dx`) (3ln(5`x` - 4) - (1/2)ln(2`x` + 1))
(guideline 4):
(1/`y`)(`dy/dx`) = (3*(1/(5`x` - 4))*5) - ((1/2)*(1/(2`x` + 1))*2)
25`x` + 19
(1/`y`)(`dy/dx`) = ------------------
(5`x` - 4)(2`x` + 1 )
(guideline 5): (remember, multiply by the given `f(x)` instead of the variable `y`)
25`x` + 19 (5`x` - 4)^{3}
`dy/dx` = ------------------ * ---------------
(5`x` - 4)(2`x` + 1 ) (2`x` + 1)^{(1/2)}
(25`x` + 19)(5`x` - 4)^{2}
`dy/dx` = -----------------
(2`x` + 1)^{(3/2)}

This answer could be checked by applying the quotient rule to `f(x)`.