Question

does anyone know where i can find the complete rules for the derivatives regarding e? I'm baffled by the first derivative of e^x...

Answer 1

you want a proof ?

Answer 2

\(\large\color{slate}{\displaystyle \frac{d}{dx}\left(e^x\right)=\lim_{h\rightarrow ~0}\frac{e^{x+h}-e^x}{h}}\)

Answer 3

use the first principles of derivatives to prove it

Answer 4

and, just to know

\(\large\color{slate}{\displaystyle \frac{d}{dx}\left(e^x\right)=e^x}\)

Answer 5

ohhhhhhhhhhhh

Answer 6

which would not be exactly true with a different exponent

Answer 7

thanks so much, wow this setup rules

Answer 8

I dunno if this is strictly "legal" but \(\dfrac{d}{dx} a^x = a^x ln(a)\)

Answer 9

yep, bibby, now i remember.... i must have been looking at the wrong chapter in my notes

Answer 10

or, if you are assuming that given that d/dx (1/x)=ln(x)
then

\(\large\color{slate}{\displaystyle y=e^x}\)
\(\large\color{slate}{\displaystyle \ln(y)=\ln(e^x)}\)
\(\large\color{slate}{\displaystyle \ln(y)=x\ln(e)}\)
\(\large\color{slate}{\displaystyle \ln(y)=x}\)
\(\large\color{slate}{\displaystyle y'~(1/y)=1}\)
\(\large\color{slate}{\displaystyle y'=1\times y}\)
\(\large\color{slate}{\displaystyle y'= y}\)\
and we know from the beginning y=e^x, so
\(\large\color{slate}{\displaystyle y'=e^x}\)

Answer 11

I mean given that d/dx lnx = 1/x the other way around... sorry

Answer 12

which means \(\dfrac{d}{dx}e^x=e^x*\ln(e)=e^x*1=e^x\)

Answer 13

When you take the Taylor series expansion of e^x you get
$$
1+x+x^2/2+x^3/6+x^4/24+x^5/120+x^6/720+\\
x^7/5040+x^8/40320+x^9/362880+x^{10}/3628800+O(x^{11})
$$
If you take the derivative of e^x, you can take the term by term derivative of the above, and again you get the Taylor Series expansion of e^x, which shows that the derivative of e^x is e^x

Answer 14

yes, you would just be getting previosu term, and there are infinity of those

Answer 15

very good approximation of e^x would it be.

Answer 16

anyway...

Answer 17

have fun, and keep in mind log. differentiation, if you ever forget or want to prove quotient or product rule for differentiating a function

Answer 18

Oh, interesting excessive
try showing it

\(\large\color{slate}{\displaystyle \frac{d^n}{dx^n}\left(xe^x\right)=ne^x+xe^x}\)

Answer 19

exercise (spelling check... )

Answer 20

can you use induction?

Answer 21

well, I am talking about just taking derivative twice or 3 times and showing that it will be like this for any nth derivative

Answer 22

yeah, that's induction pretty much

Answer 23

as I remember it

Answer 24

also derivative of cos(x) and sin(x) would be interesting.

Answer 25

I mean if you use the first prinicples.
all that matters is practice.... practice as much as you can and you will master it!

Answer 26

woah, so is this Taylor expansion thing how you find an equation for which you have the derivative?

Answer 27

$$
{d\over dx}\left (1+x+x^2/2+x^3/6+x^4/24+x^5/120+x^6/720+\\
x^7/5040+x^8/40320+x^9/362880+x^{10}/3628800+O(x^{11})\right )\\
=1+x+x^2/2+x^3/6+x^4/24+x^5/120+x^6/720+\\
x^7/5040+x^8/40320+x^9/362880+x^{10}/3628800+O(x^{11})
$$