@dageustice
f(x)=0 also has itself as its derivative, but no one seems to care at all ☹️
@dospiir8045
her : i can change him him :
@lillii9119
By the definition of derivative, we find that (a^x)' = (a^x)(a^h - 1)/h when h -> 0, Let e be the number so that (e^h - 1)/h = 1, e^h - 1 = h <=> e^h = 1 + h <=> e = (1+h)^(1/h) By definition, this is the only real number that, as the base of an exponential function, gives that function itself as a derivative
@littleantukins4415
e^x is the goat of all functions
@gilly4487
I hate my Calc 1 class rn but this video genuinely helped me. Thank you.
@kitspapp
i remember my maths professor showed us this proof in intro calc, my mind was blown
@virushk
NICE
@pagoluharshavardhangowdme1423
It also has another interesting property. Although it's slope increases exponentially it's tangents intercept on y axis increases linearly with x i.e the intercept of tangent at x=1 is 0, at x=2 is 1, at x=3 is 2.... And so on
@Umlaut95
Thank you! This helps with comprehending calculus
@loohooi6545
In addition,the difinite integaral(the area below the curve) from negative infinity to any x value of that point is also exactly to its y-coordinates.
@skyzm212
Another intereating method is using taylor's expansion series to prove that for all integer n since e^x=1+x+x^2/2+...x^n/n .then it's derivative would remain the same neglecting the rest since we're doing a derivative
@elnetini
The next step would be to show the area under the curve, which is the same value as the slope at any given point
@omerutkuerzengin3061
So nice explanation.
@bijipeter1471
Thank you,sir
@thefirsttrillionaire2925
Wow I’ll never forget this now 😮
@logosking2848
Not by coincidence, but by definition
@jesusnoagervasini8207
f(x)=0: Look at what they have to do to mimic a fraction of my power
@shadowshibe5962
Oh my God I so needed this
@RunningOnEmpty-f7r
It is quite easy to understand once it's visualused like this, but personally I really like the other explanation of its consistency which involves Taylor series. You see, e^x can be expressed as an infinite sum of x^n/n! so it goes like this: 1+x + x²/2!... and so on.... So if you find a derivative of this you'll get the exact same thing, and I really love it
@sathmika