Intuition The big idea (WHY this matters)Most functions get messier when you differentiate them (x 3 → 3 x 2 x^3 \to 3x^2 x 3 → 3 x 2 sin → cos \sin \to \cos sin → cos e x e^x e x one function that is its own derivative : its slope at every point equals its height. That single fact is why e e e e x e^x e x what is the slope of an exponential, and why is e e e
Definition Derivative from first principlesFor any f f f f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h . f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. f ′ ( x ) = lim h → 0 h f ( x + h ) − f ( x ) . slope of the tangent : rise over run as the run h h h 0 0 0
Apply this to f ( x ) = a x f(x)=a^x f ( x ) = a x a > 0 a>0 a > 0 HOW: use the exponent law a x + h = a x a h a^{x+h}=a^x\,a^h a x + h = a x a h
f ′ ( x ) = lim h → 0 a x + h − a x h = lim h → 0 a x a h − a x h = a x ⋅ lim h → 0 a h − 1 h ⏟ call this M ( a ) . f'(x)=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}
=\lim_{h\to 0}\frac{a^x a^h-a^x}{h}
=a^x\cdot\underbrace{\lim_{h\to 0}\frac{a^h-1}{h}}_{\text{call this }M(a)}. f ′ ( x ) = h → 0 lim h a x + h − a x = h → 0 lim h a x a h − a x = a x ⋅ call this M ( a ) h → 0 lim h a h − 1 .
Intuition WHY the limit factors outa x a^x a x x x x a x a^x a x h → 0 h\to 0 h → 0 h h h M ( a ) = lim h → 0 a h − 1 h M(a)=\lim_{h\to0}\frac{a^h-1}{h} M ( a ) = lim h → 0 h a h − 1 pure number depending on a a a a x a^x a x at x = 0 x=0 x = 0 (the height there is a 0 = 1 a^0=1 a 0 = 1
So for every exponential:
d d x a x = M ( a ) a x , M ( a ) = lim h → 0 a h − 1 h \boxed{\;\frac{d}{dx}a^x = M(a)\,a^x,\qquad M(a)=\lim_{h\to0}\frac{a^h-1}{h}\;} d x d a x = M ( a ) a x , M ( a ) = h → 0 lim h a h − 1 The derivative is the function back again, just scaled by the constant M ( a ) M(a) M ( a )
e e e want a base where the slope-at-0 0 0 1 1 1 its own derivative with no annoying constant out front. Define e e e that base:
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.
This is one clean way to define e ≈ 2.71828 e\approx 2.71828 e ≈ 2.71828
d d x e x = e x \boxed{\;\frac{d}{dx}e^x=e^x\;} d x d e x = e x
M ( a ) = 1 M(a)=1 M ( a ) = 1 a = e a=e a = e d d x e x = M ( e ) e x = 1 ⋅ e x = e x \frac{d}{dx}e^x = M(e)\,e^x = 1\cdot e^x = e^x d x d e x = M ( e ) e x = 1 ⋅ e x = e x Why this step? Because we chose e e e M ( e ) = 1 M(e)=1 M ( e ) = 1
Consistency check (forecast-then-verify): does this limit definition agree with the famous e = lim n → ∞ ( 1 + 1 n ) n e=\lim_{n\to\infty}(1+\tfrac1n)^n e = lim n → ∞ ( 1 + n 1 ) n h h h e h − 1 ≈ h e^h-1\approx h e h − 1 ≈ h e h ≈ 1 + h e^h\approx 1+h e h ≈ 1 + h e ≈ ( 1 + h ) 1 / h e\approx(1+h)^{1/h} e ≈ ( 1 + h ) 1/ h h = 1 / n → 0 h=1/n\to0 h = 1/ n → 0 e = lim ( 1 + 1 / n ) n e=\lim(1+1/n)^n e = lim ( 1 + 1/ n ) n
We don't want to leave M ( a ) M(a) M ( a ) HOW: write any base in terms of e e e
a = e ln a ⟹ a x = ( e ln a ) x = e ( ln a ) x . a = e^{\ln a}\quad\Longrightarrow\quad a^x=\big(e^{\ln a}\big)^x=e^{(\ln a)\,x}. a = e l n a ⟹ a x = ( e l n a ) x = e ( l n a ) x .
Now differentiate e ( ln a ) x e^{(\ln a)x} e ( l n a ) x e u e^u e u u = ( ln a ) x u=(\ln a)x u = ( ln a ) x
d d x a x = d d x e ( ln a ) x = e ( ln a ) x ⋅ ( ln a ) = a x ln a . \frac{d}{dx}a^x=\frac{d}{dx}e^{(\ln a)x}
= e^{(\ln a)x}\cdot(\ln a)
= a^x\ln a. d x d a x = d x d e ( l n a ) x = e ( l n a ) x ⋅ ( ln a ) = a x ln a .
ln a \ln a ln a a = e a=e a = e ln a = ln e = 1 \ln a=\ln e=1 ln a = ln e = 1 = 1 =1 = 1 a > e a>e a > e 0 0 0 ln a > 1 \ln a>1 ln a > 1 a < 1 a<1 a < 1 ln a < 0 \ln a<0 ln a < 0
d d x 3 x \dfrac{d}{dx}\,3^x d x d 3 x = 3 x ln 3 =3^x\ln 3 = 3 x ln 3 Why? Direct use of d d x a x = a x ln a \frac{d}{dx}a^x=a^x\ln a d x d a x = a x ln a a = 3 a=3 a = 3
d d x e 5 x \dfrac{d}{dx}\,e^{5x} d x d e 5 x u = 5 x u=5x u = 5 x u ′ = 5 u'=5 u ′ = 5 e 5 x ⋅ 5 = 5 e 5 x e^{5x}\cdot 5 = 5e^{5x} e 5 x ⋅ 5 = 5 e 5 x Why this step? e x e^x e x 5 5 5
d d x 2 x 2 \dfrac{d}{dx}\,2^{x^2} d x d 2 x 2 2 x 2 = e x 2 ln 2 2^{x^2}=e^{x^2\ln 2} 2 x 2 = e x 2 l n 2 = e x 2 ln 2 ⋅ ( 2 x ln 2 ) = 2 x 2 ( 2 x ) ln 2 =e^{x^2\ln2}\cdot(2x\ln2)=2^{x^2}\,(2x)\ln 2 = e x 2 l n 2 ⋅ ( 2 x ln 2 ) = 2 x 2 ( 2 x ) ln 2 Why? Convert to base e e e x 2 ln 2 x^2\ln2 x 2 ln 2 2 x ln 2 2x\ln2 2 x ln 2
Worked example (d) Slope ofe x e^x e x x = 0 x=0 x = 0 2 x 2^x 2 x x = 0 x=0 x = 0 e x e^x e x = e 0 = 1 =e^0=1 = e 0 = 1 2 x \quad 2^x 2 x = 2 0 ln 2 = ln 2 ≈ 0.693 =2^0\ln2=\ln2\approx0.693 = 2 0 ln 2 = ln 2 ≈ 0.693 Why it matters: 2 x 2^x 2 x slower at the start than e x e^x e x
d d x a x = x a x − 1 \dfrac{d}{dx}a^x = x\,a^{x-1} d x d a x = x a x − 1 Why it feels right: it copies 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 Why it's wrong: in x n x^n x n base varies and the exponent is fixed; in a x a^x a x exponent varies and the base is fixed — opposite situation, so the power rule does not apply.
Fix: exponential variable → a x ln a a^x\ln a a x ln a
d d x e x = x e x − 1 \dfrac{d}{dx}e^x = xe^{x-1} d x d e x = x e x − 1 a = e a=e a = e Fix: e x e^x e x
d d x e 5 x = e 5 x \dfrac{d}{dx}e^{5x}=e^{5x} d x d e 5 x = e 5 x Why it feels right: "e stuff e^{\text{stuff}} e stuff = x =x = x Fix: multiply by the inner derivative: e 5 x ⋅ 5 e^{5x}\cdot5 e 5 x ⋅ 5
M ( a ) = log 10 a M(a)=\log_{10} a M ( a ) = log 10 a Why it feels right: "log of a a a Fix: the proof forces a = e ln a a=e^{\ln a} a = e l n a natural log. It must be ln a \ln a ln a log 10 a \log_{10}a log 10 a
Recall Feynman: explain to a 12-year-oldImagine money that grows. A normal curve like x 2 x^2 x 2 different shape. e x e^x e x how steep it is right now is exactly how tall it is right now . So if it's 5 5 5 5 5 5 2 x 2^x 2 x 3 x 3^x 3 x natural log of the base. e e e 1 1 1
"e e e ln \ln ln
e x e^x e x a x a^x a x ln a \ln a ln a a x ln a a^x\ln a a x ln a
What is d d x e x \frac{d}{dx}e^x d x d e x e x e^x e x (it is its own derivative).
What is d d x a x \frac{d}{dx}a^x d x d a x Why does a x a^x a x a x M ( a ) a^x\,M(a) a x M ( a ) Because
a x + h = a x a h a^{x+h}=a^x a^h a x + h = a x a h , and
a x a^x a x is constant in
h h h so it leaves the limit;
M ( a ) = lim h → 0 a h − 1 h M(a)=\lim_{h\to0}\frac{a^h-1}{h} M ( a ) = lim h → 0 h a h − 1 .
What does the limit M ( a ) = lim h → 0 a h − 1 h M(a)=\lim_{h\to0}\frac{a^h-1}{h} M ( a ) = lim h → 0 h a h − 1 ln a \ln a ln a ; it is the slope of
a x a^x a x at
x = 0 x=0 x = 0 .
How is e e e e e e is the base with
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 .
Derive d d x a x \frac{d}{dx}a^x d x d a x e e e a x = e ( ln a ) x a^x=e^{(\ln a)x} a x = e ( l n a ) x ; chain rule gives
e ( ln a ) x ⋅ ln a = a x ln a e^{(\ln a)x}\cdot\ln a=a^x\ln a e ( l n a ) x ⋅ ln a = a x ln a .
What is d d x e 5 x \frac{d}{dx}e^{5x} d x d e 5 x e 5 x e^{5x} e 5 x 5 e 5 x 5e^{5x} 5 e 5 x ; chain rule multiplies by inner derivative
5 5 5 .
Differentiate 2 x 2 2^{x^2} 2 x 2 2 x 2 ( 2 x ) ln 2 2^{x^2}(2x)\ln 2 2 x 2 ( 2 x ) ln 2 .
Why is "d d x a x = x a x − 1 \frac{d}{dx}a^x=xa^{x-1} d x d a x = x a x − 1 Power rule is for fixed exponent/varying base; here the exponent varies, so it doesn't apply.
Slope of 2 x 2^x 2 x x = 0 x=0 x = 0 ln 2 ≈ 0.693 \ln 2\approx0.693 ln 2 ≈ 0.693 .
M of a = limit of a^h - 1 over h
e = limit of 1+1/n to the n
Intuition Hinglish mein samjhoDekho, idea simple hai: jab tum kisi exponential a x a^x a x a x + h = a x a h a^{x+h}=a^x a^h a x + h = a x a h a x a^x a x M ( a ) = lim h → 0 a h − 1 h M(a)=\lim_{h\to0}\frac{a^h-1}{h} M ( a ) = lim h → 0 h a h − 1 x = 0 x=0 x = 0 M ( a ) M(a) M ( a )
Ab e e e M ( e ) = 1 M(e)=1 M ( e ) = 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
Baaki bases ke liye trick: kisi bhi a a a a = e ln a a=e^{\ln a} a = e l n a a x = e ( ln a ) x a^x=e^{(\ln a)x} a x = e ( l n a ) x ln a \ln a ln a d d x a x = a x ln a \frac{d}{dx}a^x=a^x\ln a d x d a x = a x ln a M ( a ) M(a) M ( a ) ln a \ln a ln a
Sabse common galti: power rule (x n → n x n − 1 x^n\to nx^{n-1} x n → n x n − 1 e 5 x e^{5x} e 5 x 5 e 5 x 5e^{5x} 5 e 5 x e 5 x e^{5x} e 5 x
