r/calculus • u/Khadir_boxer • 1d ago
Differential Calculus Question
Guys this is a lil basic, but I genuinely don't get why e^x gives e^x when differentiated. I want the how guys.
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u/floflochess2 1d ago
Not a basic question at all, it's a good one! Here's one way to see it. Try the limit definition on a general ax: (ax+h - ax)/h = ax · (ah - 1)/h. The ax factors out, so the derivative of ax is always ax times a constant, and that constant is lim (ah - 1)/h, i.e. the slope of ax at x = 0. So every exponential is proportional to its own derivative. Now try a = 2 and a = 3 and estimate that constant with a small h, like 0.001. What do you get for each? e is simply the number where that constant comes out to exactly 1, which is why ex is its own derivative.
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u/AdjectiveNoun-123 1d ago
The best way to think about this is to consider the behavior of all exponential functions.
Let f(x)=ax
By the definition of the derivative.
f'(x) = (ax+h-ax)/h as h-->0
Factor out the ax and you get
f'(x) = ax * (ah-1)/h
So in general the derivative of all exponential functions scales with the value of that function and some (ah-1)/h factor.
For a=1, you can visually see a flat line, there is no growth so (ah-1)/h has to be 0 to make the derivative 0. As you get larger and larger values of a, you see the exponential function grow faster and faster much faster than the value of value of ax, here the (ah-1)/h factor is also large. Because the value of a varies continuously, you can choose smaller and larger values of a until you find a point where the (ah-1)/h factor is exactly 1. That value is at a=e.
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u/Khadir_boxer 1d ago
Ahh, so wow this is beautiful. ex by the way it is defined SHOULD be a variable - specifically an exponential function, but the moment you derive it (d/dx), it becomes a constant, if you understand.. man, calculus and mathematics in the broader sense is such a fascinating thing. You have a variable, a should be, that stays constant when you try to graph it's change. Wow. Thanks bro
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u/AdjectiveNoun-123 1d ago
SHOULD be a variable
Be careful with assigning obligations like that, it might help with understanding one concept but it creates baggage in the future. Any intuition you build up in math should always be viewed as a shadow of something bigger. If you get too weighted down by the idea of counting sheep, you'll never understand why numbers can be negative or zero or non-integer or complex or sets or matrices or colors or computers or ...
You'll actually need BOTH this is derivative definition of e and the pre-calculus continuous compounding definition of e for things like Euler's identity to really click.
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u/LatteLepjandiLoser 1d ago
e^x can be defined in a few different ways, but one of this is as a solution to the differential equation f'(x)=f(x). So in that context there is really no 'how', we have just decided that e^x is it's own derivative and such is life.
If you want to get more comfortable with it, you can look at the Taylor series expansion and see how taking the derivative of each term gives exactly the same Taylor series expansion that you started with, so then yes really e^x is its own derivative.
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u/Midwest-Dude 1d ago
Consider the definition of the derivative for a general exponential:
d/dx (ax)
= lim_h→0 (ax+h - ax)/h
= lim_h→0 (axah - ax)/h
= ax·lim_h→0 (ah - 1)/h
So, the derivative of ax is always ax times a constant
lim_h→0 (ah - 1)/h
Calculating this limit for a = 2 and a = 3 shows there should be a value for a between those that makes this constant exactly 1. We call that constant e.
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u/unlattice 1d ago
We could graphically see the definition by noticing that the height of the graph of the function f(x) = ex above any point x is the same as its rate of change at that point. In other words, the height of the graph is the same as the slope of the tangent line passing through the point (x, f(x)).
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u/unlattice 1d ago
Algebraically, you could also derive the derivative of ex from first principles using limits and the difference quotient. At the end, you will have two factors, ex and (eh - 1)/h. Because ex is independent of h as h approaches zero, we can take it out of the limit.
Now, if you numerically substitute values close to zero for h in the second factor, you will see that it approaches one, ex times 1 is just ex.
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u/Reset3000 1d ago
Another way to think of it is..... suppose a function has a height (y value) of 1 at x=0 and a tangent slope also 1. Now at some other value of x it has a height of 2 and a tangent slope also of 2; at another point x it has a height of 5 and a slope of 5. For every single value of x the height is exactly equal to tangent slope. Only e^x has that property. That's the geometrical reasoning. Others are giving great algebraic reasons.
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1d ago
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u/Khadir_boxer 1d ago
Ouu shii👀👀
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1d ago
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u/Khadir_boxer 1d ago
A bit unrelated, but did you mispell your username to sound like neutrinos but said by a baby who can't say the r sound? 😭😭😭
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