proooooooooooooooove

[am-banawi]

prove that
d/dx (e^x)=e^x

[caesar]

السلام عليكم :

تفضل يا سيدي :

[عبدالله عبادي]

[am-banawi]

الشكر الجزيل لكما
اخي قيصر واخي عبدالله

[am-banawi]

هل هناك طريقة للاثبات بالتعريف

[am-banawi]

?????????????????????????????????

[عبدالله عبادي]

[Ould Youbba]

السلام عليكم ورحمة الله وبركاته.

prove that
d/dx (e^x)=e^x

لتكن .
هذه الدالة هي الدالة العكسية للدالة : المعرفة من نحو .
وبما أن الدالة قابلة للاشتقاق على وتحقق :

فإن (والتي هي الدالة العكسية لـ ) قابلة للاشتقاق على ولدينا لكل عدد حقيقي :

تحياتي.

[salah ahmad]

Derivative of e^x by Limit Definition

Date: 05/21/2002 at 22:40:18
From: Jeff King
Subject: derivative of e^x by limit definition

Dear Dr. Math,

I've tried (as has my entire calculus class) to prove that the
derivative of e^x is e^x by the limit definition of the derivative
(without using the Taylor expansion for e^x) and we cannot seem to
get past one last step.

In setting up the form of the limit definition of the derivative we've
made e(x+delta x) into e^x*e^(delta x) for convenience and left the
-e^x in the numerator of the limit definition. Then by factoring we
determined that the limit as delta x goes to 0 is

(e^x(e^(delta x)-1))/delta x.

The e^x alone in the numerator would seem to be our best bet, but we
can't make the rest of the function go to 1 as some people have
guessed it should.

How would one prove the limit definition or just that

(e^(delta x)-1)/(delta x) = 1

as x tends to infinity? Thanks for your assistance!

Date: 05/29/2002 at 22:41:49
From: Doctor Fwg
Subject: Re: derivative of e^x by limit definition

Dear Jeff,

Here is a possible solution that does not involve the use of a Taylor
expansion. I hope the notation I have used here will be easy enough
to follow.

Definition of e is:

1) e = Limit (1 + 1/n)^n
n->Inf

However, if n = 1/h, then

2) e = Limit (1 + h)^(1/h)
h->0

If f(x) = e^x, the definition of f'(x) is:

3) f'(x) = Limit [f(x + h) - f(x)]/h
h->0

But:

4) f(x + h) = e^(x+h) = (e^x)(e^h)

So:

5) f'(x) = Limit [(e^x)(e^h) - e^x]/h
h->0

= Limit [(e^x)((e^h) - 1)]/h
h->0

But, raising both sides of equation (2) to the power of h yields:

6) e^h = Limit [(1 + h)^(1/h)]^h
h->0

= Limit [(1+ h)]
h->0

Placing this value of e^h into equation (5) yields:

7) f'(x) = Limit [(e^x){(1 + h) - 1}]/h
h->0

= Limit [e^x(h/h)]
h->0

But, h/h = 1, so:

8) f'(x) = Limit [e^x(1)]
h->0

= e^x

With Best Regards,

Doctor Fwg, The Math Forum

[salah ahmad]

Derivative of e^x by Limit Definition

Date: 05/21/2002 at 22:40:18
From: Jeff King
Subject: derivative of e^x by limit definition

Dear Dr. Math,

I've tried (as has my entire calculus class) to prove that the
derivative of e^x is e^x by the limit definition of the derivative
(without using the Taylor expansion for e^x) and we cannot seem to
get past one last step.

In setting up the form of the limit definition of the derivative we've
made e(x+delta x) into e^x*e^(delta x) for convenience and left the
-e^x in the numerator of the limit definition. Then by factoring we
determined that the limit as delta x goes to 0 is

(e^x(e^(delta x)-1))/delta x.

The e^x alone in the numerator would seem to be our best bet, but we
can't make the rest of the function go to 1 as some people have
guessed it should.

How would one prove the limit definition or just that

(e^(delta x)-1)/(delta x) = 1

as x tends to infinity? Thanks for your assistance!

Date: 05/29/2002 at 22:41:49
From: Doctor Fwg
Subject: Re: derivative of e^x by limit definition

Dear Jeff,

Here is a possible solution that does not involve the use of a Taylor
expansion. I hope the notation I have used here will be easy enough
to follow.

Definition of e is:

1) e = Limit (1 + 1/n)^n
n->Inf

However, if n = 1/h, then

2) e = Limit (1 + h)^(1/h)
h->0

If f(x) = e^x, the definition of f'(x) is:

3) f'(x) = Limit [f(x + h) - f(x)]/h
h->0

But:

4) f(x + h) = e^(x+h) = (e^x)(e^h)

So:

5) f'(x) = Limit [(e^x)(e^h) - e^x]/h
h->0

= Limit [(e^x)((e^h) - 1)]/h
h->0

But, raising both sides of equation (2) to the power of h yields:

6) e^h = Limit [(1 + h)^(1/h)]^h
h->0

= Limit [(1+ h)]
h->0

Placing this value of e^h into equation (5) yields:

7) f'(x) = Limit [(e^x){(1 + h) - 1}]/h
h->0

= Limit [e^x(h/h)]
h->0

But, h/h = 1, so:

8) f'(x) = Limit [e^x(1)]
h->0

= e^x

With Best Regards,

Doctor Fwg, The Math Forum