# Logaritmic diffrentation restrictions

1. Nov 1, 2013

### new_at_math

Why does if a function can equal zero, since you can't take the log of zero.

EDIT:
let me rephrase the question; where can logarithmic differentiation not be used?

Last edited: Nov 2, 2013
2. Nov 1, 2013

### UltrafastPED

Your question is not clear. Give an example.

3. Nov 2, 2013

### HallsofIvy

Staff Emeritus
"Logarithmic differentiation" cannot be used where the logarithm does not exist! Since logarithm functions have domain "all positive numbers" that would be where the function you are taking the logarithm of is not positive.

You could, however, deal with the absolute value of the function, using the fact that the derivative of |f(x)| is the same as the the derivative of f if f(x) is positive, the negative of the derivative of f if f(x) is negative, and undefined if f(x)= 0.

4. Nov 2, 2013

### new_at_math

?

let me get this right:

lets say I have f(x) = x^3
then I can log both sides since f(x) is positive,even though it has values that are both 0 and negative

after, log., differentiating I get 3x^2

so I can always use logarithmic differentiation whenever the equation is not equal to zero, since having a negative sign on one side can be fixed by taking the absolute value of both sides. the fact that the function contains, in other words: can output, negative or zero values does not matter.

5. Nov 3, 2013

### Staff: Mentor

???
f(x) = x3 is positive only when x > 0.
As already noted, when x ≤ 0, log(x) is not defined.

6. Nov 4, 2013

### HallsofIvy

Staff Emeritus
Then ln(f)= ln(x^3)= 3 ln(x) as long as x is positive. f'(x)/f(x)= f'(x)/x^3= 3/x so f'(x)= 3x^2.

If x is negative, neither ln(x^3) nor ln(x) exists but we can take g(x)= |x^3|= -f(x) or f(x)= -g(x) and say that f'(x)= -g'(x). For x negative, g(x)= -x^2 is positive so we can say that g'(x)=-3x^2 and then f'(x)= -(-3x^2)= 3x^2. Derivatives are not necessarily continuous but they do satisfy the "intermediate value theorem- for x between a and b. f'(x) must be between f(a) and f(b). In particular, since 3x^2 and -3x^2 go to 0 as x goes to 0, f'(0), if it exists, must be 0.

Of course. this is the hard way to do this derivative!