anonymous
  • anonymous
What is the derivative of e^(2x) I am confused because the derivative e^x is e^x. But they say you need to use the chain rule. Which is f'(g(x))*g'(x). I have used that on exponents but in this I just get, 2x * e^(2x-1) *e Which is not the answer =/
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
2e^(2x)
amistre64
  • amistre64
\[y=e^{2x}\] \[ln(y)=2x\] \[y'/y = 2\] \[y'=2y;\ but\ y=e^{2x}\] \[y' = 2e^{2x}\]
anonymous
  • anonymous
So I just have to use logs, to differentiate it.

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Shayaan_Mustafa
  • Shayaan_Mustafa
@amistre64 will you kindly elaborate elaborate your 3rd step i.e. y'/y=2 how did you get this? please.
s3a
  • s3a
(Chain) rule: d/dx e^[f(x)] = df(x)/dx * e^[f(x)] Examples: 1) d/dx e^x = dx/dx * e^x = 1 * e^x = e^x 2) d/dx e^(2x) = d(2x)/dx * e^(2x) = 2 * e^(2x)
s3a
  • s3a
Shayaan_Mustafa, if you treat y as a function (and not just a regular variable) then you take the derivative of y as if it's just a regular variable and you turn the y into a y'.
s3a
  • s3a
Example if y = y(x), d/dx (y^2) = 2y * y'
anonymous
  • anonymous
1) d/dx e^x = dx/dx * e^x = 1 * e^x = e^x why does e^x = 1?
anonymous
  • anonymous
o wait
s3a
  • s3a
d(x)/dx = 1
s3a
  • s3a
in this case f(x) = x.
s3a
  • s3a
This being e^x.
anonymous
  • anonymous
ok , I kinda get it. So all I have to do is use the chain rule, why did he use the ln rule above?
s3a
  • s3a
He was proving it.
s3a
  • s3a
In other words, the very first time mathematicians had to do this, they had to justify why what I said works.
s3a
  • s3a
I think.
s3a
  • s3a
If someone can confirm, that'd be great (for me).
anonymous
  • anonymous
ok lol =P
s3a
  • s3a
:)
anonymous
  • anonymous
Thanks for the help everyone, I'm a little closer to understanding it.

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