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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 744

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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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

hi iamaditya

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 744

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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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

hi iamaditya

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

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 744

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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