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## #1 2017-08-31 02:46:46

Member
From: Planet Mars
Registered: 2016-11-15
Posts: 707

### Plzz help me in this proof

Hi all;

I was trying to prove that the derivative of e^x is e^x, but I couldn't.

Now, how to proceed further?

Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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## #2 2017-08-31 07:40:24

bob bundy
Registered: 2010-06-20
Posts: 8,108

### Re: Plzz help me in this proof

Firstly note that
(whoops ... cannot find the Latex for lim as delta x tends to zero

lim as delta x tends to zero

is the gradient of e^x at (0,1)

What do you know about e^x ?

If you know it's a power series then you can get the result easily by differentiating that.

I usually start a class with what follows where the class no nothing yet about e^x

Consider y = a^x  for some real a

At x = 0 y = 1.  As x gets bigger y gets bigger and as x gets more negative y get smaller (but still positive)

Thus we have a 'family' of curves all with similar properties.  If y = b^x is another where b is slightly bigger than a, then b^x is bigger than a^x for positive x and smaller for negative x.

As a may be any real number the 'family' all go through (0,1) all have a fixed but not yet known gradient at (0,1) ; let's say that gradient is k ; and dy/dx = k.a^x

Thus dy/dx = ky for all a in the family.

[note: if a is negative then y = -a^x is the familiar member of the family and just the mirror image of y = |a|^x in the y axis.  So we only need consider a to be zero or positive]

If a = 0 the 'curve' is just the line y=1.

As a gets bigger from 0, curves have steadily increasing gradient at (0,1) so choose the one where this gradient, k,
is
1 as special and call that value of a, the special letter e.

So dy/dx = y for this value  e.

So the function has the property that it differentiates to give itself.

Are there other functions with this property?

Let's suppose there are two, df/dx = f and dg/dx = g.

Consider f/g and differentiate using the quotient rule

So if we integrate this we get

So all functions with the property are multiples of e^x

Now consider the power series

This will differentiate to give itself so it must be a multiple of e^x.

But the series has value 1 at x = 0 so it must be e^x itself.

That's as far as I'll go for now.  I can continue to show that integral of 1/x is ln(x) and hence obtain a value for k for all values of a.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #3 2017-08-31 21:12:47

Member
From: Planet Mars
Registered: 2016-11-15
Posts: 707

### Re: Plzz help me in this proof

Hmm nice, but I took some time to grasp it.

Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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## #4 2017-09-02 23:31:23

bob bundy
Registered: 2010-06-20
Posts: 8,108

### Re: Plzz help me in this proof

Don't worry; that's about three hour lessons for an A level class.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #5 2017-09-03 02:52:21

Member
From: Planet Mars
Registered: 2016-11-15
Posts: 707

### Re: Plzz help me in this proof

Hmm I see. I am trying to learn differential calculus now since its easy for beginners.

Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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