Intuition The one function that is its own rate of changeImagine money in a bank where the interest is added continuously , and the interest rate equals the current amount. The bigger the pile, the faster it grows — and the speed of growth is exactly the size of the pile. That self-feeding process is e x e^x e x slope at every point equals its height . No other function (except its scalings) does this.
Definition Natural exponential functionThe natural exponential function is f ( x ) = e x f(x)=e^x f ( x ) = e x e ≈ = = 2.71828 = = e\approx ==2.71828== e ≈== 2.71828 == a x a^x a x = = 1 = = ==1== == 1 == y y y x = 0 x=0 x = 0
WHY does a special base matter?
For a general a x a^x a x d d x a x = a x ⋅ k \dfrac{d}{dx}a^x = a^x\cdot k d x d a x = a x ⋅ k k k k a a a want the base for which k = 1 k=1 k = 1 e e e
We start from the definition of the derivative (first principles):
d d x a x = lim h → 0 a x + h − a x h \frac{d}{dx}a^x = \lim_{h\to 0}\frac{a^{x+h}-a^x}{h} d x d a x = lim h → 0 h a x + h − a x
Why this step? The derivative is by definition the limit of average rate of change; we are not allowed to assume any rule yet.
Factor out a x a^x a x h h h
= a x lim h → 0 a h − 1 h = a^x\lim_{h\to 0}\frac{a^{h}-1}{h} = a x lim h → 0 h a h − 1
Why this step? a x + h = a x ⋅ a h a^{x+h}=a^x\cdot a^h a x + h = a x ⋅ a h a x a^x a x h h h
Call that limit k ( a ) = lim h → 0 a h − 1 h k(a)=\displaystyle\lim_{h\to 0}\frac{a^h-1}{h} k ( a ) = h → 0 lim h a h − 1
d d x a x = k ( a ) a x \frac{d}{dx}a^x = k(a)\,a^x d x d a x = k ( a ) a x
Why this matters: the shape of the derivative is the same function back , just scaled by k ( a ) k(a) k ( a )
Numerically check the limit: for a = 2 a=2 a = 2 k ≈ 0.693 k\approx 0.693 k ≈ 0.693 a = 3 a=3 a = 3 k ≈ 1.099 k\approx 1.099 k ≈ 1.099 k = 1 k=1 k = 1 a = e ≈ 2.718 a=e\approx 2.718 a = e ≈ 2.718
Feature
Value
WHY
Passes through
( 0 , 1 ) (0,1) ( 0 , 1 ) e 0 = 1 e^0=1 e 0 = 1 0 = 1 ^0=1 0 = 1
Passes through
( 1 , e ) ≈ ( 1 , 2.72 ) (1,e)\approx(1,2.72) ( 1 , e ) ≈ ( 1 , 2.72 ) definition of e e e
Range
y > 0 y>0 y > 0 e x e^x e x
Horizontal asymptote
y = 0 y=0 y = 0 x → − ∞ x\to-\infty x → − ∞ e − x = 1 / e x → 0 e^{-x}=1/e^x\to 0 e − x = 1/ e x → 0
Behaviour as x → + ∞ x\to+\infty x → + ∞
→ + ∞ \to+\infty → + ∞ grows without bound, faster than any polynomial
Slope at ( 0 , 1 ) (0,1) ( 0 , 1 )
= 1 =1 = 1 the defining property
Slope at any ( x , e x ) (x,e^x) ( x , e x )
= e x =e^x = e x d d x e x = e x \frac{d}{dx}e^x=e^x d x d e x = e x
Concavity
always concave up
d 2 d x 2 e x = e x > 0 \frac{d^2}{dx^2}e^x=e^x>0 d x 2 d 2 e x = e x > 0
Intuition Slope = height, visuallyAt the point ( 1 , e ) (1,e) ( 1 , e ) ≈ 2.72 \approx 2.72 ≈ 2.72 also ≈ 2.72 \approx 2.72 ≈ 2.72
Worked example 4 — Forecast-then-verifyForecast: Is the slope of y = e x y=e^x y = e x x = 2 x=2 x = 2 x = 1 x=1 x = 1 Reason: slope = height, and height at x = 2 x=2 x = 2 e 2 ≈ 7.39 e^2\approx7.39 e 2 ≈ 7.39 x = 1 x=1 x = 1 e ≈ 2.72 e\approx2.72 e ≈ 2.72 Verify: slope( 2 ) = e 2 ≈ 7.39 (2)=e^2\approx7.39 ( 2 ) = e 2 ≈ 7.39 ( 1 ) = e ≈ 2.72 (1)=e\approx2.72 ( 1 ) = e ≈ 2.72
d d x e x = x e x − 1 \frac{d}{dx}e^x = xe^{x-1} d x d e x = x e x − 1 Why it feels right: the power rule d d x x n = n x n − 1 \frac{d}{dx}x^n=nx^{n-1} d x d x n = n x n − 1 The fix: the power rule needs the variable in the base and a constant in the exponent . Here the variable is in the exponent — that's a different animal. Correct: d d x e x = e x \frac{d}{dx}e^x=e^x d x d e x = e x
e x e^x e x x x x Why it feels right: negative x x x sounds like it should push y y y The fix: e − x = 1 e x e^{-x}=\dfrac{1}{e^x} e − x = e x 1 y = 0 y=0 y = 0
e e e 2.7 2.7 2.7 Why it feels right: 2.7 2.7 2.7 The fix: e = 2.71828 … e=2.71828\ldots e = 2.71828 … irrational ; it's defined by the slope-1 property, not by any decimal.
Recall Feynman: explain to a 12-year-oldThink of a magic snowball rolling downhill. The bigger it gets, the faster it grows — and its speed is exactly how big it already is. e x e^x e x high the curve is tells you exactly how steep it is. And the number e ≈ 2.72 e\approx 2.72 e ≈ 2.72
"E is Effortlessly its own Echo" — differentiate e x e^x e x echo it right back: e x e^x e x slope = height for e x e^x e x
What is the derivative of e x e^x e x e x e^x e x (it is its own derivative)
What defines the base e e e a x a^x a x The base whose graph has slope exactly
1 1 1 at
x = 0 x=0 x = 0 , i.e.
lim h → 0 e h − 1 h = 1 \lim_{h\to0}\frac{e^h-1}{h}=1 lim h → 0 h e h − 1 = 1 Approximate value of e e e 2.71828 … 2.71828\ldots 2.71828 … (irrational)
What point does y = e x y=e^x y = e x ( 0 , 1 ) (0,1) ( 0 , 1 ) with slope
1 1 1 Range of e x e^x e x y > 0 y>0 y > 0 (all positive reals)
Horizontal asymptote of e x e^x e x y = 0 y=0 y = 0 as
x → − ∞ x\to-\infty x → − ∞ Why isn't d d x e x = x e x − 1 \frac{d}{dx}e^x = xe^{x-1} d x d e x = x e x − 1 Power rule needs a variable base & constant exponent; here the variable is in the exponent
Tangent line to e x e^x e x x = 0 x=0 x = 0 Slope of e x e^x e x ( x , e x ) (x,e^x) ( x , e x ) e x e^x e x — it equals the height
Derivative of e 3 x e^{3x} e 3 x 3 e 3 x 3e^{3x} 3 e 3 x (chain rule)
Is e x e^x e x Concave up, since second derivative
e x > 0 e^x>0 e x > 0 differentiate from first principles
second derivative positive
Limit k a equals lim of a^h-1 over h
Euler number e approx 2.71828
Continuous growth process
Intuition Hinglish mein samjhoDekho, e x e^x e x slope har point pe uski height ke barabar hota hai. Matlab jahan curve ki value 2.72 2.72 2.72 2.72 2.72 2.72 e ≈ 2.718 e\approx 2.718 e ≈ 2.718 e x e^x e x
First principles se derive karein toh d d x a x = a x ⋅ lim h → 0 a h − 1 h \frac{d}{dx}a^x = a^x \cdot \lim_{h\to 0}\frac{a^h-1}{h} d x d a x = a x ⋅ lim h → 0 h a h − 1 k k k a a a k = 1 k=1 k = 1 e e e d d x e x = e x \frac{d}{dx}e^x = e^x d x d e x = e x
Graph yaad rakho: ( 0 , 1 ) (0,1) ( 0 , 1 ) y > 0 y>0 y > 0 y = 0 y=0 y = 0 power rule mat lagao — d d x e x \frac{d}{dx}e^x d x d e x x e x − 1 xe^{x-1} x e x − 1
