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anonymous
 5 years ago
In Lecture 6, Mr. David Jerison talks about natural logs, and presents to us:
w=ln(x) so e^w=x, (d/dx)e^w=(d/dx)x=1, so [(d/dx)e^w](dw/dx)=1
This implies (dw/dx)=1. How so, and how did he even arrive at this conclusion?
anonymous
 5 years ago
In Lecture 6, Mr. David Jerison talks about natural logs, and presents to us: w=ln(x) so e^w=x, (d/dx)e^w=(d/dx)x=1, so [(d/dx)e^w](dw/dx)=1 This implies (dw/dx)=1. How so, and how did he even arrive at this conclusion?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You missed a step. He simplified the (d/dx)e^w to simply e^w. It went something like this: w=lnx e^w=x (d/dx)e^w=(d/dx)x So here we know we have to chain rule the e^w. e^w(dw/dx)=1 (dw/dx)=1/(e^w) Just substitute back the x (dw/dx)=1/x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The way he wrote it was a bit weird for me too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh so is it kind of like \[(dx/dx)e^w*(dw/dx) \] and the (dx/dx) cancels out?
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