Regarding logarithmic differentiation

[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'.

I was wondering if I could go about this in another way.
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.

 

[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)##.

 

[Permanence]

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)##.

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!