Question

(d/dx)e^(lny) = dy/dx,... why does this not equal e^(lny)(dy/dx) by chain rule?

Answer 1

why does it not equal (y)(dy/dx)

Answer 2

set u = ln y
du = dy/y

y = e^ln y = e^ln u
d(e^lny)/dx = (d(e^u)/du)(du/dx) = y. d(e^u) / dy = y(dy/y)/dx = dy/dx


note: d(e^u) = d(e^lny) = dy

Answer 3

y(dy/dx)=dy/dx??

Answer 4

http://www.wolframalpha.com/input/?i=y%28dy%2Fdx%29

Answer 5

\(\huge x^{\log_x a}=a\)

Answer 6

so, \(\huge e^{\ln y}=y\)

Answer 7

d/dx (e^{ln y}) = d/dx (y) = dy/dx

Answer 8

yeah, but still the chain rule should apply in the same way, that it assumes the dirivative of x^2 = 2x*(d/dx(x))= 2x(1)= 2x

Answer 9

it should be that (d/dx)e^(lny)=e^(lny)*dy/dx = y(dy/dx)

Answer 10

(*assuming you are using the chain rule definition, which i understand no reason to say that you couldn't)

Answer 11

applying chain rule
\(\large d/dx [e^{\ln y}]=e^{\ln y}d/dx[\ln y] = e^{\ln y}(1/y)d/dx[y]=e^{\ln y}(1/y)dy/dx\)

Answer 12

(d/dx)e^(lny)=e^(lny)*dy/dx <------NO
(d/dx)e^(lny)=e^(lny)*d[ln y]/dx

Answer 13

\(\large d/dx [e^{\ln y}]=e^{\ln y}d/dx[\ln y] \\ = e^{\ln y}(1/y)d/dx[y]=e^{\ln y}(1/y)dy/dx \\ =(y)(1/y)dy/dx = dy/dx\)

Answer 14

you get same answer using chain rule also.

Answer 15

ok the foundations of mathematics still hold, for now ... XD, i see where i have gone wrong thank you for your explination! :)

Answer 16

welcome ^_^