I looked up how to solve this d/dx of lne^x and found this on yahoo answers:
(ln(e^x))'=
e^x'/e^x=
e^x/e^x=
1
Can someone verify the correctness of this person's steps and explain the prerequisite/intermediate steps involved. I'm having trouble seeing how it became a fraction in the first place. Perhaps its algebraic log rule that I don't know?
(ln(e^x))'=
e^x'/e^x=
e^x/e^x=
1
Can someone verify the correctness of this person's steps and explain the prerequisite/intermediate steps involved. I'm having trouble seeing how it became a fraction in the first place. Perhaps its algebraic log rule that I don't know?
-
the derivative of the natural logarithm of anything is 1/x
meaning (ln(e^x))'=1/e^x
but you also have to apply the chain rule for the inside function which is e^x and since the derivative of e^x is just e^x then,
(ln(e^x))'=(1/e^x)*e^x
=e^x/e^x
they cancel out and you end up with =1
meaning (ln(e^x))'=1/e^x
but you also have to apply the chain rule for the inside function which is e^x and since the derivative of e^x is just e^x then,
(ln(e^x))'=(1/e^x)*e^x
=e^x/e^x
they cancel out and you end up with =1
-
it became a fraction because it is (ln (something with x in it))
and, as always the derivative, where I use d/dx (whatever) to mean the derivative of (whatever)
d/dx (ln (something with x in it)) = {d/dx ( something with x in it) } / (something with x in it)
e.g.
ln (x^2) = {d/dx (x^2)} / x^2 = (2x) / x^2 = 2 / x
in the example you quoted from Y!A:
d/dx ln(e^x) = { d/dx (e^x) } / e^x = e^x / e^x = 1
As the first poster suggested, the way that answer did it was the long way. It is easier to notice that ln( e^{anything} = anything because ln and e are inverses of each other. The ln undoes the e.
so
ln (e^x) = x
thus
d/dx ln(e^x) = d/dx (x) = 1
It is not an algebratic (?) log rule, it is just a derivative. I think it is dangerous to call it a log rule, it encourages memorization rather than understanding. It is only a "rule" if you do not know how to differentiate in some sense. I get what you mean, I just kind of cringe when students talk about if something they missed in their understand was "just" some "rule" or something, when what they really missed was how to differentiate properly in general.
and, as always the derivative, where I use d/dx (whatever) to mean the derivative of (whatever)
d/dx (ln (something with x in it)) = {d/dx ( something with x in it) } / (something with x in it)
e.g.
ln (x^2) = {d/dx (x^2)} / x^2 = (2x) / x^2 = 2 / x
in the example you quoted from Y!A:
d/dx ln(e^x) = { d/dx (e^x) } / e^x = e^x / e^x = 1
As the first poster suggested, the way that answer did it was the long way. It is easier to notice that ln( e^{anything} = anything because ln and e are inverses of each other. The ln undoes the e.
so
ln (e^x) = x
thus
d/dx ln(e^x) = d/dx (x) = 1
It is not an algebratic (?) log rule, it is just a derivative. I think it is dangerous to call it a log rule, it encourages memorization rather than understanding. It is only a "rule" if you do not know how to differentiate in some sense. I get what you mean, I just kind of cringe when students talk about if something they missed in their understand was "just" some "rule" or something, when what they really missed was how to differentiate properly in general.