Quick question regarding mathematical constant e

[15357]

e defined as lim|x to inf| (1 + 1/x)^x

I've been told to prove that the derivative of e^x or (e^x)' = e^x

So I went ahead by using the rule for the derivative of inverse functions in which {|inv|f(x)}' = 1/f'(x)


and as f(x) = ln x, the derivative f'(x) = 1/x

because f'(x) = lim|h to 0| [{f(x+h) - f(x)}/h]

and therefore = lim (ln (x+h) - ln(x))/h

= lim (ln (1 + h/x))/h

and to help, defining z as x/h

= lim [1/h * ln(1+1/z)]

= lim [z/x * ln(1+1/z)]

= lim [1/x * ln(1+1/z)^z]

= 1/x ln lim (1+1/z)^z

the latter part in which it is defined as e as stated above,

so = 1/x * ln e

and as ln e = 1 because log|e|e = 1

therefore = 1/x


so again, f(x) = ln x, the derivative f'(x) = 1/x

and using this, and the fact that the inv. function of e^x is ln x

(ln x)' = 1/x = 1/(e^x)'


which left me confused since obviously x does not equal e^x which is supposed to be the derivative (e^x)'


Did I do something wrong here?

 

[ríomhaire]

Have you tried proving it from first principles?

 

[15357]

Well no, I suppose I can copy the proof from the textbook, provided I can find it, but I'm still intrigued about this thing.

I don't think I did anything wrong here, but the answer doesn;t come up right..

 

[Mr Stabby]

have you tried expanding them as taylor's series. I'm too busy to think about it but i think you want some form of expansions.

 

[99.vikram]

By definition, derivative of e^x:

= lt dx -> 0: (e^(x+dx) - e^x) / ((x + dx) - x)
= lt dx -> 0: (e^x*e^dx - e^x) / dx
= e^x * lt dx -> 0: (e^dx - 1) / dx

Now, using the power series definition of e:

= e^x * lt dx -> 0: ( (1 + dx + dx^2/2! + dx^3/3! + dx^4/4! + ...) - 1) / dx
= e^x * lt dx -> 0: (dx + dx^2/2! + dx^3/3! + ...)/dx
= e^x * lt dx -> 0: 1 + dx/2! + dx^2/3! + ...
= e^x

Hence proved, ab initio (from first principles.)

EDIT: MrStabby is right, Taylor expansion also works to give the same result.

 

[DEATHMASTER]

I like how you call this a quick question.

 

[99.vikram]

(ln x)' = 1/x = 1/(e^x)'


which left me confused since obviously x does not equal e^x which is supposed to be the derivative (e^x)'


Did I do something wrong here?

Yes. You tried to do a variable substitution in one step and screwed it up (I've done it before.)

(ln x)' = 1/x

^This part is true
But what does that little apostrophe mean?

d(ln x)/dx = 1/x

Now, replace all x above by e^x:

d(ln e^x)/d(e^x) = 1/e^x
=> dx/d(e^x) = 1/e^x

Now invert the above:
=> d(e^x)/dx = e^x

TA DAA!!!

 

[15357]

Ohhhh so I got the inverse wrong.

Thanks guys, esp. Vikram who answered my question of "What did I do wrong" most directly

 

[JUL3]

Numbers, your numbers make my head hurt.

 

[<RJMC>]

x is your mom

 

[ríomhaire]

(ln x)' = 1/x

^This part is true
But what does that little apostrophe mean?

Differentiation.
F' = dF/dx

 

[mindless_moder]

Differentiation.
F' = dF/dx

He knows what it means , that was for Number's benefit.