Derivative of Exponential Proof

Derivative of Exponential Proof

Category:

Education

Tags:

• math
• mathematics
• proof
• calculus
• derivative
• shahz
• pollardrho06

[StepBack13]

I agree with Monty whole-heartedly.

This isn't a rigourous proof.

By saying that lim((e^(x+dx)-1)/dx)=1, you have used the definition of e^x as being it's own derivative in your proof of this same fact.

[pollardrho06]

OK. I will look into it and come up with a rigorous one... thanks for commenting...

[fcasson]

Nicely done.

[pollardrho06]

Thanks for watching. Appreciate it.

[StepBack13]

@AidanLod This proof is incomplete, as the final step sets the following limit as equal to one, without proof:

lim(Delta-x ->0) (exp(Delta-x)-1)/(Delta-x)

This limit approaches a form of 0/0, since the numerator gives (exp(0)-1) as Delta-x approaches 0. Thus the numerator and denominator go to 0. Hence, the limit can't be evaluated as is. We must use L'Hopital's rule. But that forces us to differentiate the top and bottom... and we can't do this, since the derivative of exp(x) is unknown.

[clh161]

the lim delta x ->0 (exp(delta x) -1)/(delta x) just magically turns to 1,why?

[MOHSLM]

@wallyroyal

But I think the derivation of taylors series for e^x depends on the fact that we already know the derivative of e^x which isn't true.

[MOHSLM]

I don't really understand how the derivation is incomplete. well if u understand what does it mean to evaluate a limit u will know that the limit is concerned with what happens as we approach zero, it doesn't care about what happens when the values is exactly zero. so now we can use this fact to evaluate the lim as Delta x approaches zero from the left hand by substituting a value smaller than zero and very close to it and the for the right hand side a bigger value. Man nicely done !

[wallyroyal]

Use the two term taylor series for e^delta (or triangle x, i prefer delta) 2 lines from the end. It gives 1+delta-1 all over delta which cancels and leaves e^x. Sorry if this has been mentioned before

[StepBack13]

@AidanLod Here is a proof I conducted that does not 'beg the question': youtube.com/watch?v=IOivKRokYp­k

[AidanLod]

@StepBack13 this is the choice i would use to proof that e^x's derivative is equal to its original function.

[AidanLod]

you have a flaw somewhere.

[AidanLod]

but then if lim

hx -> 0 then e^hx is = to 1, so 1-1 = 0 therefore 0/0 is undefinable?