Question

log derivatives: y = e^-x *lnx

Answer 1

i did product rule i got -e^-x *lnx + e^-x *1/x

Answer 2

i second that answer ^

Answer 3

Simplest form:
\[e^{-x}\left(\frac{1}{x} -\ln{x}\right)\]

Answer 4

Do you know implicit differentiation?

Answer 5

yea

Answer 6

No need for implicit differentiation here

Answer 7

well the answer simplified is -e^-x *lnx+1/x... if i did product rule correctly, what happened to the e^-x?

Answer 8

if you factor out hero's answer, the negative is in the minus sign

Answer 9

@Hero There is another solution.

We know that e^-x *lnx = x^(-x)

So d/dx y = d/dx x^(-x)

It's kind of difficult to do it this way, so before derivative both sides, put ln to both sides, like that:
lny = lnx^(-x)
lny = -xlnx

Now derivative both sides. (HINT: lnu = u'/u)

Answer 10

i just want to know what happens to the e^-x

Answer 11

will it be just x?

Answer 12

or lnx?

Answer 13

@swin2013 wait, so lnx is not part of exponent or what?

Answer 14

no it's product rule. e^-x is multiplied by lnx

Answer 15

Well, that changes the problem. Yeah, just use product rule.

Are you asking what will happen to e^(-x) if we derivative it?

Answer 16

no, the second part of the product rule is e^-x + 1/x. The answer is -e^-x*lnx + 1/x. I see how there is a -e^-x *lnx but i do not see where the e^-x went

Answer 17

I think you did something wrong. Not sure exactly what you mean by "the second part of the product rule is e^-x + 1/x."

Let f(x) be e^(-x) and g(x) be lnx\[\huge \frac{d}{dx} f(x)g(x) = f'(x)·g(x) + f(x)·g'(x)\]In this way,you should get this, the simplest form: \[\huge -\frac{\ln x}{e^x} + \frac{1}{x e^x}\] Which can be factored to this:\[\huge e^{-x}\left( -\ln x + \frac{1}{x} \right)\]

Is this clear? Does this answer your question?

Answer 18

i meant after the addition sign of the whole product rule

Answer 19

I get it lol

Answer 20

Well, it's\[\frac{e^{-x}}{x}\]not\[e^{-x} + \frac{1}{x}\]

Answer 21

oops i meant a multiplication sign.

Answer 22

Okay, In second part of product rule, we didn't derivative e^(-x) so thats where it came from. Does this answer your question?