Intuition The big pictureZoom into the graph of a smooth function near a point and it looks like a straight line . That line is the tangent line . So if you know one exact value f ( a ) f(a) f ( a ) f ′ ( a ) f'(a) f ′ ( a ) estimate nearby values without recomputing the messy function. Linear approximation is just "ride the tangent line a little way."
Definition Linear approximation (linearization)The linearization of f f f x = a x=a x = a a a a L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L(x) = f(a) + f'(a)\,(x-a) L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) x x x a a a f ( x ) ≈ L ( x ) f(x) \approx L(x) f ( x ) ≈ L ( x ) L L L L ine.
Let y = f ( x ) y=f(x) y = f ( x ) differential d x dx d x x x x y y y d y = f ′ ( x ) d x dy = f'(x)\,dx d y = f ′ ( x ) d x d y dy d y along the tangent line when x x x d x dx d x y y y Δ y = f ( x + d x ) − f ( x ) \Delta y = f(x+dx)-f(x) Δ y = f ( x + d x ) − f ( x ) Δ y ≈ d y \Delta y \approx dy Δ y ≈ d y
The derivative is defined as a limit:
f ′ ( a ) = lim x → a f ( x ) − f ( a ) x − a . f'(a) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a}. f ′ ( a ) = lim x → a x − a f ( x ) − f ( a ) .
"Limit equals f ′ ( a ) f'(a) f ′ ( a ) x x x a a a close to f ′ ( a ) f'(a) f ′ ( a ) ε ( x ) \varepsilon(x) ε ( x ) → 0 \to 0 → 0 x → a x\to a x → a f ( x ) − f ( a ) x − a = f ′ ( a ) + ε ( x ) . \frac{f(x)-f(a)}{x-a} = f'(a) + \varepsilon(x). x − a f ( x ) − f ( a ) = f ′ ( a ) + ε ( x ) .
Multiply both sides by ( x − a ) (x-a) ( x − a ) f ( x ) − f ( a ) = f ′ ( a ) ( x − a ) + ε ( x ) ( x − a ) . f(x)-f(a) = f'(a)(x-a) + \varepsilon(x)(x-a). f ( x ) − f ( a ) = f ′ ( a ) ( x − a ) + ε ( x ) ( x − a ) .
So
f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) ⏟ L ( x ) + ε ( x ) ( x − a ) ⏟ error . f(x) = \underbrace{f(a)+f'(a)(x-a)}_{L(x)} + \underbrace{\varepsilon(x)(x-a)}_{\text{error}}. f ( x ) = L ( x ) f ( a ) + f ′ ( a ) ( x − a ) + error ε ( x ) ( x − a ) .
Why this is great: the error is ε ( x ) ⋅ ( x − a ) \varepsilon(x)\cdot(x-a) ε ( x ) ⋅ ( x − a ) small number times a small number . It shrinks faster than ( x − a ) (x-a) ( x − a ) a a a L ( x ) L(x) L ( x ) f f f
Pick a base point a a a near your target where f ( a ) f(a) f ( a ) f ′ ( a ) f'(a) f ′ ( a ) easy/exact .Compute f ( a ) f(a) f ( a ) f ′ ( a ) f'(a) f ′ ( a )
Plug into L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L(x)=f(a)+f'(a)(x-a) L ( x ) = f ( a ) + f ′ ( a ) ( x − a )
Evaluate at your target x x x d x = x − a dx = x-a d x = x − a f ( x ) ≈ f ( a ) + d y f(x)\approx f(a)+dy f ( x ) ≈ f ( a ) + d y
4.1 \sqrt{4.1} 4.1 Step 1 — choose f f f a a a Let f ( x ) = x f(x)=\sqrt{x} f ( x ) = x a = 4 a=4 a = 4 Why? 4 = 2 \sqrt 4=2 4 = 2 4.1 4.1 4.1 4 4 4 Step 2 — slope. f ′ ( x ) = 1 2 x f'(x)=\tfrac{1}{2\sqrt x} f ′ ( x ) = 2 x 1 f ′ ( 4 ) = 1 4 f'(4)=\tfrac1{4} f ′ ( 4 ) = 4 1 Why? The tangent slope controls the line.
Step 3 — line. L ( x ) = 2 + 1 4 ( x − 4 ) L(x)=2+\tfrac14(x-4) L ( x ) = 2 + 4 1 ( x − 4 ) Step 4 — evaluate. L ( 4.1 ) = 2 + 1 4 ( 0.1 ) = 2.025 L(4.1)=2+\tfrac14(0.1)=2.025 L ( 4.1 ) = 2 + 4 1 ( 0.1 ) = 2.025 4.1 = 2.02485 … \sqrt{4.1}=2.02485\ldots 4.1 = 2.02485 … 0.0001 0.0001 0.0001
( 1.02 ) 10 (1.02)^{10} ( 1.02 ) 10 f ( x ) = x 10 f(x)=x^{10} f ( x ) = x 10 a = 1 a=1 a = 1 d x = 0.02 dx=0.02 d x = 0.02 Why a = 1 a=1 a = 1 1 10 = 1 1^{10}=1 1 10 = 1 f ′ ( x ) = 10 x 9 f'(x)=10x^9 f ′ ( x ) = 10 x 9 f ′ ( 1 ) = 10 f'(1)=10 f ′ ( 1 ) = 10 d y = f ′ ( 1 ) d x = 10 ( 0.02 ) = 0.2 dy=f'(1)\,dx = 10(0.02)=0.2 d y = f ′ ( 1 ) d x = 10 ( 0.02 ) = 0.2 ( 1.02 ) 10 ≈ f ( 1 ) + d y = 1 + 0.2 = 1.2 (1.02)^{10}\approx f(1)+dy = 1+0.2 = 1.2 ( 1.02 ) 10 ≈ f ( 1 ) + d y = 1 + 0.2 = 1.2 ≈ 1.219 \approx 1.219 ≈ 1.219 Why close? d x dx d x
≈ 1 + n x \approx 1+nx ≈ 1 + n x f ( x ) = ( 1 + x ) n f(x)=(1+x)^n f ( x ) = ( 1 + x ) n a = 0 a=0 a = 0 f ( 0 ) = 1 f(0)=1 f ( 0 ) = 1 f ′ ( 0 ) = n ( 1 + 0 ) n − 1 = n f'(0)=n(1+0)^{n-1}=n f ′ ( 0 ) = n ( 1 + 0 ) n − 1 = n ( 1 + x ) n ≈ 1 + n x ( x small ) . (1+x)^n \approx 1+nx \quad (x\text{ small}). ( 1 + x ) n ≈ 1 + n x ( x small ) . Why this step? Setting a = 0 a=0 a = 0 ( x − a ) (x-a) ( x − a ) x x x sin x ≈ x \sin x\approx x sin x ≈ x e x ≈ 1 + x e^x\approx 1+x e x ≈ 1 + x ln ( 1 + x ) ≈ x \ln(1+x)\approx x ln ( 1 + x ) ≈ x 0 0 0
Worked example Error propagation (the practical payoff)A sphere's radius is measured as r = 10 r=10 r = 10 d r = 0.05 dr=0.05 d r = 0.05 V = 4 3 π r 3 V=\tfrac43\pi r^3 V = 3 4 π r 3 d V = 4 π r 2 d r = 4 π ( 100 ) ( 0.05 ) = 20 π ≈ 62.8 cm 3 dV = 4\pi r^2\,dr = 4\pi(100)(0.05)=20\pi\approx 62.8\ \text{cm}^3 d V = 4 π r 2 d r = 4 π ( 100 ) ( 0.05 ) = 20 π ≈ 62.8 cm 3 Why differentials here? We don't know the true error, only that d r dr d r Δ V ≈ d V \Delta V\approx dV Δ V ≈ d V d V V = 3 d r r = 3 ( 0.005 ) = 1.5 % \tfrac{dV}{V}=3\,\tfrac{dr}{r}=3(0.005)=1.5\% V d V = 3 r d r = 3 ( 0.005 ) = 1.5%
Common mistake "Just use the originalf f f Why it feels right: you can plug into f f f 4.1 \sqrt{4.1} 4.1 ( 1.02 ) 10 (1.02)^{10} ( 1.02 ) 10 cheap, accurate-enough estimates and error analysis , where the form d y = f ′ ( x ) d x dy=f'(x)dx d y = f ′ ( x ) d x
Common mistake Choosing a bad base pointa a a Feels right: "any a a a ( x − a ) 2 (x-a)^2 ( x − a ) 2 Fix: pick the nearest a a a f ( a ) , f ′ ( a ) f(a),f'(a) f ( a ) , f ′ ( a ) 4.1 \sqrt{4.1} 4.1 a = 4 a=4 a = 4 a = 0 a=0 a = 0
Δ y \Delta y Δ y d y dy d y Feels right: both are "the change in y y y But Δ y \Delta y Δ y true curve change; d y dy d y tangent line change. They're only approximately equal, and the gap is exactly the curvature error. Don't claim Δ y = d y \Delta y = dy Δ y = d y
Common mistake Dropping the( x − a ) (x-a) ( x − a ) f ( x ) ≈ f ( a ) + f ′ ( a ) f(x)\approx f(a)+f'(a) f ( x ) ≈ f ( a ) + f ′ ( a ) Fix: slope must be multiplied by the run ( x − a ) (x-a) ( x − a )
Recall Feynman: explain to a 12-year-oldImagine a curvy road. Right where you're standing, walk in the exact direction the road points (its slope). For a few steps you stay almost on the road, even though it slowly curves away. So instead of doing hard curvy math, you "walk straight a little." f ( a ) f(a) f ( a ) f ′ ( a ) f'(a) f ′ ( a ) ( x − a ) (x-a) ( x − a )
Mnemonic Remember the formula"Start + Slope × Step" → f ( a ) + f ′ ( a ) ( x − a ) f(a) + f'(a)\,(x-a) f ( a ) + f ′ ( a ) ( x − a ) d y = f ′ ( x ) d x dy=f'(x)\,dx d y = f ′ ( x ) d x d d d d ash along the tangent."
What is the linearization of f f f a a a L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L(x)=f(a)+f'(a)(x-a) L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) Why does linear approximation work (one line)? The limit definition forces
f ( x ) = L ( x ) + ε ( x ) ( x − a ) f(x)=L(x)+\varepsilon(x)(x-a) f ( x ) = L ( x ) + ε ( x ) ( x − a ) with error shrinking faster than
( x − a ) (x-a) ( x − a ) .
Define the differential d y dy d y d y = f ′ ( x ) d x dy=f'(x)\,dx d y = f ′ ( x ) d x , the rise along the tangent line for an input change
d x dx d x .
Difference between Δ y \Delta y Δ y d y dy d y Δ y \Delta y Δ y = true change of the curve;
d y dy d y = change along the tangent line;
Δ y ≈ d y \Delta y\approx dy Δ y ≈ d y .
Approximate ( 1 + x ) n (1+x)^n ( 1 + x ) n x x x 1 + n x 1+nx 1 + n x (linearize at
a = 0 a=0 a = 0 ).
How do you pick the base point a a a Nearest value to the target where
f ( a ) f(a) f ( a ) and
f ′ ( a ) f'(a) f ′ ( a ) are exact/easy.
Estimate 4.1 \sqrt{4.1} 4.1 a = 4 a=4 a = 4 2 + 1 4 ( 0.1 ) = 2.025 2+\tfrac14(0.1)=2.025 2 + 4 1 ( 0.1 ) = 2.025 .
Relative error in volume of a sphere from radius error? d V V = 3 d r r \tfrac{dV}{V}=3\,\tfrac{dr}{r} V d V = 3 r d r .
sin x \sin x sin x e x e^x e x ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) 0 0 0 sin x ≈ x \sin x\approx x sin x ≈ x ,
e x ≈ 1 + x e^x\approx 1+x e x ≈ 1 + x ,
ln ( 1 + x ) ≈ x \ln(1+x)\approx x ln ( 1 + x ) ≈ x .
Derivative as a limit — the definition that is the proof here.Tangent line — the geometric object L ( x ) L(x) L ( x ) Taylor series — linear approximation is the degree-1 Taylor polynomial; next terms cut the error.Newton's method — repeatedly linearizing to solve f ( x ) = 0 f(x)=0 f ( x ) = 0 Error propagation — physics/lab use of d y = f ′ ( x ) d x dy=f'(x)\,dx d y = f ′ ( x ) d x Concavity and second derivative — controls whether L L L
Difference quotient plus error e(x)
Linearization L(x)=f(a)+f'(a)(x-a)
Intuition Hinglish mein samjhoDekho, idea bahut simple hai. Koi bhi smooth curve agar tum kisi point ke bilkul paas zoom karo, toh woh ek straight line jaisi dikhne lagti hai — woh line hai tangent line . Toh agar tumhe ek point pe exact value f ( a ) f(a) f ( a ) f ′ ( a ) f'(a) f ′ ( a ) f ( x ) ≈ f ( a ) + f ′ ( a ) ( x − a ) f(x)\approx f(a)+f'(a)(x-a) f ( x ) ≈ f ( a ) + f ′ ( a ) ( x − a )
Yeh kaam kyun karta hai? Kyunki derivative ki definition hi limit se aati hai: paas mein difference quotient f ′ ( a ) f'(a) f ′ ( a ) ε ( x ) ( x − a ) \varepsilon(x)(x-a) ε ( x ) ( x − a ) ( x − a ) (x-a) ( x − a ) a a a
Differentials bas isi ka chhota bhai hai: d y = f ′ ( x ) d x dy=f'(x)\,dx d y = f ′ ( x ) d x d x dx d x d y dy d y Δ y \Delta y Δ y Δ y ≈ d y \Delta y\approx dy Δ y ≈ d y error propagation . Jaise sphere ka radius 10 10 10 0.05 0.05 0.05 d V = 4 π r 2 d r dV=4\pi r^2\,dr d V = 4 π r 2 d r d V / V = 3 d r / r dV/V = 3\,dr/r d V / V = 3 d r / r
Exam tip: base point a a a sabse paas waala easy value choose karo (jaise 4.1 \sqrt{4.1} 4.1 a = 4 a=4 a = 4 ( 1 + x ) n ≈ 1 + n x (1+x)^n\approx 1+nx ( 1 + x ) n ≈ 1 + n x sin x ≈ x \sin x\approx x sin x ≈ x e x ≈ 1 + x e^x\approx 1+x e x ≈ 1 + x x x x