# Regarding logarithmic differentiation

1. Nov 3, 2013

### Permanence

Thank you for viewing my thread. I have been given the following steps for logarithmic differentiation:
1. Take natural logarithms of both sides of an equation y = f(x) and use the Laws of Logarithms to simplify.
2. Differentiate implicitly with respect to x.
3. Solve the resulting equation for y'.

1. Take natural logarithms of both sides of an equation y = f(x) and use the Laws of Logarithms to simplify.
2. At this point I would have something like:
ln(y) = a+b+c
Instead of doing implicit differentiation , could I do this:
e^[ln(y)] = e^(a+b+c)
y = e^(a+b+c)

Thanks.

Last edited: Nov 4, 2013
2. Nov 3, 2013

### fzero

The problem with this is that $e^{a+b+c}$ is just equal to $f(x)$. The laws of logarithms that you use to simplify $\ln f(x)$ don't make $f(x)$ any simpler than using the rules for simplifying exponents and products. The reason to take the log is to make differentiating things like $(x+2)^x$ a bit simpler, since it is easier to differentiate $x \ln (x+2)$.

3. Nov 4, 2013

### Permanence

Yeah I get what you're saying now. I completely forgot a step lol. For some reason the first time I looked at it I forgot that I was doing differentiation. In my mind I had it as:
y' = e^(a+b+c)

Thank you!