Why is the derivative of e^x also e^x?
Can someone describe it in words? i.e. without a mathematical proof?
I am 100% sure it's right, keith. You don't use the power rule for this. Awms: thanks.
Can someone describe it in words? i.e. without a mathematical proof?
I am 100% sure it's right, keith. You don't use the power rule for this. Awms: thanks.
?
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Keith is wrong; the derivative of e^x is indeed e^x.
I can only offer a mathematical answer. It is because e^x is DEFINED to be the series
1+x+x^2/2! + x^3/3! + ...
Now if you just differentiate term by term you find that
d/dx ( 1+x+x^2/2! + x^3/3! + ...) = 1+x+x^2/2! + x^3/3! + ..
In other words, the derivative of e^x is e^x.
The definition of e^x as a series and theorems which say it is OK to differentiate this series term by term are from a part of mathematics known as power series.
mano
Im not really sure how to say it but i think the easiest way to see is to use the taylor expansion of e^x=1+x+x^2/2!+x^3/3!+x^4/4!+....
clearly then if you differentiate e^x by differentiating its taylor series term by term you end with what you started with. Sorry I cant think of any better way of saying it!
Awms A
Because the y-coordinate of y = e^x gives the slope of the tangent line?
Sounds a little like it's begging the question, doesn't it? Really, this is a phenomenon of mathematics, so you'll have to use mathematics (e.g. proof) to show why it is.
keith
It's been a long time for me, but I don't think that's correct.
If y = e^x, then
dy/dx = xe^(x-1)