Differentiation is a linear operation . That single word packs everything in this note: the derivative of a combination of functions is just the same combination of their derivatives. You never have to go back to the limit definition for a sum like x 2 + sin x x^2 + \sin x x 2 + sin x
WHY does this matter? Almost every function you meet is built by adding, subtracting, and scaling simpler pieces. If derivatives respected this building process, calculus becomes modular. They do — and these three rules are the proof.
Definition The three linearity rulesLet f f f g g g x x x c c c
Sum rule: d d x [ f ( x ) + g ( x ) ] = = = f ′ ( x ) + g ′ ( x ) = = \dfrac{d}{dx}\big[f(x)+g(x)\big] = ==f'(x)+g'(x)== d x d [ f ( x ) + g ( x ) ] === f ′ ( x ) + g ′ ( x ) == Difference rule: d d x [ f ( x ) − g ( x ) ] = = = f ′ ( x ) − g ′ ( x ) = = \dfrac{d}{dx}\big[f(x)-g(x)\big] = ==f'(x)-g'(x)== d x d [ f ( x ) − g ( x ) ] === f ′ ( x ) − g ′ ( x ) == Constant multiple rule: d d x [ c f ( x ) ] = = = c f ′ ( x ) = = \dfrac{d}{dx}\big[c\,f(x)\big] = ==c\,f'(x)== d x d [ c f ( x ) ] === c f ′ ( x ) ==
Combined into one linearity statement:
d d x [ a f ( x ) + b g ( x ) ] = a f ′ ( x ) + b g ′ ( x ) \frac{d}{dx}\big[a\,f(x)+b\,g(x)\big] = a\,f'(x)+b\,g'(x) d x d [ a f ( x ) + b g ( x ) ] = a f ′ ( x ) + b g ′ ( x )
WHAT we start from. The only thing we are allowed to assume is the definition of the derivative:
h ′ ( x ) = lim Δ x → 0 h ( x + Δ x ) − h ( x ) Δ x h'(x)=\lim_{\Delta x\to 0}\frac{h(x+\Delta x)-h(x)}{\Delta x} h ′ ( x ) = lim Δ x → 0 Δ x h ( x + Δ x ) − h ( x ) limit laws : the limit of a sum is the sum of limits, and constants pull out of limits.
Let h ( x ) = f ( x ) + g ( x ) h(x)=f(x)+g(x) h ( x ) = f ( x ) + g ( x )
h ′ ( x ) = lim Δ x → 0 [ f ( x + Δ x ) + g ( x + Δ x ) ] − [ f ( x ) + g ( x ) ] Δ x h'(x)=\lim_{\Delta x\to 0}\frac{\big[f(x+\Delta x)+g(x+\Delta x)\big]-\big[f(x)+g(x)\big]}{\Delta x} h ′ ( x ) = lim Δ x → 0 Δ x [ f ( x + Δ x ) + g ( x + Δ x ) ] − [ f ( x ) + g ( x ) ]
Why this step? I just substituted h = f + g h=f+g h = f + g
Now regroup the numerator by pairing f f f f f f
= lim Δ x → 0 [ f ( x + Δ x ) − f ( x ) Δ x + g ( x + Δ x ) − g ( x ) Δ x ] =\lim_{\Delta x\to 0}\left[\frac{f(x+\Delta x)-f(x)}{\Delta x}+\frac{g(x+\Delta x)-g(x)}{\Delta x}\right] = lim Δ x → 0 [ Δ x f ( x + Δ x ) − f ( x ) + Δ x g ( x + Δ x ) − g ( x ) ]
Why this step? Addition is commutative, so I can shuffle the four terms into two difference quotients. This is the whole trick.
Apply the sum law for limits (allowed because both limits exist by assumption):
= lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x ⏟ f ′ ( x ) + lim Δ x → 0 g ( x + Δ x ) − g ( x ) Δ x ⏟ g ′ ( x ) = f ′ ( x ) + g ′ ( x ) ■ =\underbrace{\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}}_{f'(x)}+\underbrace{\lim_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}}_{g'(x)}=f'(x)+g'(x)\;\;\blacksquare = f ′ ( x ) Δ x → 0 lim Δ x f ( x + Δ x ) − f ( x ) + g ′ ( x ) Δ x → 0 lim Δ x g ( x + Δ x ) − g ( x ) = f ′ ( x ) + g ′ ( x ) ■
Let h ( x ) = c f ( x ) h(x)=c\,f(x) h ( x ) = c f ( x )
h ′ ( x ) = lim Δ x → 0 c f ( x + Δ x ) − c f ( x ) Δ x = lim Δ x → 0 c ⋅ f ( x + Δ x ) − f ( x ) Δ x h'(x)=\lim_{\Delta x\to 0}\frac{c\,f(x+\Delta x)-c\,f(x)}{\Delta x}=\lim_{\Delta x\to 0}c\cdot\frac{f(x+\Delta x)-f(x)}{\Delta x} h ′ ( x ) = lim Δ x → 0 Δ x c f ( x + Δ x ) − c f ( x ) = lim Δ x → 0 c ⋅ Δ x f ( x + Δ x ) − f ( x )
Why this step? Factor out c c c
= c lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = c f ′ ( x ) ■ =c\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=c\,f'(x)\;\;\blacksquare = c lim Δ x → 0 Δ x f ( x + Δ x ) − f ( x ) = c f ′ ( x ) ■
Why this step? A constant factor passes through a limit — it doesn't depend on Δ x \Delta x Δ x
Write f − g = f + ( − 1 ) ⋅ g f-g = f + (-1)\cdot g f − g = f + ( − 1 ) ⋅ g sum rule , then the constant multiple rule with c = − 1 c=-1 c = − 1 d d x [ f − g ] = f ′ + ( − 1 ) g ′ = f ′ − g ′ ■ \frac{d}{dx}[f-g]=f'+(-1)g'=f'-g'\;\;\blacksquare d x d [ f − g ] = f ′ + ( − 1 ) g ′ = f ′ − g ′ ■
Intuition Why difference is "free"Subtraction is just adding a negative. Once you have linearity, you don't prove difference separately — it falls out. This is the 80/20 lesson: master the sum + constant rules; everything else is recombination.
Worked example Example 1 — a polynomialDifferentiate y = 3 x 4 − 5 x 2 + 7 y = 3x^4 - 5x^2 + 7 y = 3 x 4 − 5 x 2 + 7
y ′ = d d x ( 3 x 4 ) − d d x ( 5 x 2 ) + d d x ( 7 ) y' = \frac{d}{dx}(3x^4) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(7) y ′ = d x d ( 3 x 4 ) − d x d ( 5 x 2 ) + d x d ( 7 ) Why? Sum/difference rule splits the three terms.
= 3 ⋅ 4 x 3 − 5 ⋅ 2 x = 12 x 3 − 10 x = 3\cdot 4x^3 - 5\cdot 2x = 12x^3 - 10x = 3 ⋅ 4 x 3 − 5 ⋅ 2 x = 12 x 3 − 10 x Why? Constant multiple pulls 3 3 3 5 5 5 x n x^n x n 7 7 7 0 0 0
Worked example Example 2 — mixing function typesDifferentiate f ( x ) = 2 sin x + 4 e x − 1 3 ln x f(x) = 2\sin x + 4e^x - \tfrac{1}{3}\ln x f ( x ) = 2 sin x + 4 e x − 3 1 ln x
f ′ ( x ) = 2 cos x + 4 e x − 1 3 ⋅ 1 x f'(x) = 2\cos x + 4e^x - \frac{1}{3}\cdot\frac{1}{x} f ′ ( x ) = 2 cos x + 4 e x − 3 1 ⋅ x 1 Why? Each term is a constant times a known derivative (sin ′ = cos \sin'=\cos sin ′ = cos ( e x ) ′ = e x (e^x)'=e^x ( e x ) ′ = e x ln ′ = 1 / x \ln' = 1/x ln ′ = 1/ x
Worked example Example 3 — rewrite first, then applyDifferentiate g ( x ) = x 2 + 6 x 2 g(x) = \dfrac{x^2 + 6\sqrt{x}}{2} g ( x ) = 2 x 2 + 6 x
Step 1 — rewrite as constant multiples of powers:
g ( x ) = 1 2 x 2 + 3 x 1 / 2 g(x) = \tfrac{1}{2}x^2 + 3x^{1/2} g ( x ) = 2 1 x 2 + 3 x 1/2 Why? Dividing by 2 2 2 1 2 \tfrac12 2 1 x = x 1 / 2 \sqrt{x}=x^{1/2} x = x 1/2 Step 2 — differentiate term by term:
g ′ ( x ) = 1 2 ⋅ 2 x + 3 ⋅ 1 2 x − 1 / 2 = x + 3 2 x g'(x) = \tfrac{1}{2}\cdot 2x + 3\cdot\tfrac{1}{2}x^{-1/2} = x + \frac{3}{2\sqrt{x}} g ′ ( x ) = 2 1 ⋅ 2 x + 3 ⋅ 2 1 x − 1/2 = x + 2 x 3
Common mistake "The derivative of a product is the product of derivatives — like the sum rule."Why it feels right: The sum rule split cleanly , so the brain over-generalizes: surely multiplication splits too? Symmetry is seductive.
The fix: Linearity only covers + + + − - − × \times × product rule : ( f g ) ′ = f ′ g + f g ′ (fg)'=f'g+fg' ( f g ) ′ = f ′ g + f g ′ d d x ( x ⋅ x ) = 2 x \frac{d}{dx}(x\cdot x)=2x d x d ( x ⋅ x ) = 2 x x ′ ⋅ x ′ = 1 ⋅ 1 = 1 ≠ 2 x x'\cdot x' = 1\cdot 1 = 1 \ne 2x x ′ ⋅ x ′ = 1 ⋅ 1 = 1 = 2 x
Common mistake Forgetting that a constant's derivative is0 0 0 Why it feels right: Constants "look like" terms, so students differentiate 7 7 7 7 7 7 1 1 1 The fix: c f ( x ) c f(x) c f ( x ) f ( x ) = 1 f(x)=1 f ( x ) = 1 c ⋅ 0 = 0 c\cdot 0 = 0 c ⋅ 0 = 0 7 7 7 0 0 0 not 7 7 7
Common mistake Mishandling the sign in subtraction with a multi-term second function.Differentiating f − ( x 2 − 3 x ) f - (x^2 - 3x) f − ( x 2 − 3 x ) f ′ − 2 x − 3 f' - 2x - 3 f ′ − 2 x − 3 Wrong — the minus distributes: f ′ − ( 2 x − 3 ) = f ′ − 2 x + 3 f' - (2x - 3) = f' - 2x + 3 f ′ − ( 2 x − 3 ) = f ′ − 2 x + 3 The fix: Treat − ( g ) -(g) − ( g ) ( − 1 ) ⋅ g (-1)\cdot g ( − 1 ) ⋅ g whole g g g
Recall Predict before computingFor y = 10 x 3 − x 5 + cos x y = 10x^3 - \dfrac{x}{5} + \cos x y = 10 x 3 − 5 x + cos x forecast y ′ y' y ′ y ′ = 30 x 2 − 1 5 − sin x y' = 30x^2 - \tfrac{1}{5} - \sin x y ′ = 30 x 2 − 5 1 − sin x x 5 = 1 5 x \tfrac{x}{5}=\tfrac15 x 5 x = 5 1 x 1 5 \tfrac15 5 1 cos ′ = − sin \cos' = -\sin cos ′ = − sin
Sum rule for derivatives d d x [ f + g ] = f ′ + g ′ \frac{d}{dx}[f+g]=f'+g' d x d [ f + g ] = f ′ + g ′ Difference rule d d x [ f − g ] = f ′ − g ′ \frac{d}{dx}[f-g]=f'-g' d x d [ f − g ] = f ′ − g ′ Constant multiple rule d d x [ c f ] = c f ′ \frac{d}{dx}[cf]=c\,f' d x d [ c f ] = c f ′ Which limit law makes the sum rule work? The limit of a sum equals the sum of the limits (when both exist).
Why is the difference rule not proved separately? Because
f − g = f + ( − 1 ) g f-g=f+(-1)g f − g = f + ( − 1 ) g , so sum + constant-multiple rules give it for free.
Derivative of a constant c c c 0 0 0 (a constant has zero slope).
Does linearity apply to products? No — products need the product rule
( f g ) ′ = f ′ g + f g ′ (fg)'=f'g+fg' ( f g ) ′ = f ′ g + f g ′ .
One-line "linearity" statement d d x [ a f + b g ] = a f ′ + b g ′ \frac{d}{dx}[af+bg]=af'+bg' d x d [ a f + b g ] = a f ′ + b g ′ d d x ( 3 x 4 − 5 x 2 + 7 ) \frac{d}{dx}(3x^4-5x^2+7) d x d ( 3 x 4 − 5 x 2 + 7 ) 12 x 3 − 10 x 12x^3-10x 12 x 3 − 10 x
Recall Feynman: explain to a 12-year-oldImagine you're tracking two runners' speeds. If you add their positions together to make a "combined" position, the speed of that combined thing is just the two speeds added up. And if a runner ran on a treadmill set to triple speed, his recorded speed is just three times normal. That's all these rules say: adding stays added, scaling stays scaled. But you can't do this for "runner A's distance times runner B's distance" — multiplying mixes them, and that needs a different rule.
"Slopes add, constants ride along."
Sums → add the slopes. Scaling by c c c c c c rides along outside the derivative. Subtraction = adding a negative.
Limit laws (sum, scalar, product of limits) — the engine behind these proofs.Definition of the derivative (limit of difference quotient) — where we started.Power rule — combined with linearity to differentiate any polynomial.Product rule — the non-linear sibling; contrast carefully.Linearity of integration — the integral version of the exact same idea.Linear operators — the abstract structure: d d x \frac{d}{dx} d x d
Derivative limit definition
Limit laws: sum splits, constants pull out
Constant multiple rule c f'
Linearity of differentiation
Modular calculus of built-up functions
Intuition Hinglish mein samjhoDekho, yeh teen rules ka matlab sirf ek hai: differentiation ek linear kaam hai. Iska seedha matlab — agar function ko aap tukdo me jodte ho (f + g f+g f + g f ′ + g ′ f'+g' f ′ + g ′ c c c c c c ( c f ) ′ = c f ′ (cf)' = c f' ( c f ) ′ = c f ′ f − g f-g f − g f + ( − 1 ) g f + (-1)g f + ( − 1 ) g
Yeh kaam kyun karta hai? Kyunki limit ki property hi aisi hai: do cheezon ke sum ki limit, unki limits ka sum hoti hai, aur constant limit ke bahar nikal aata hai. Hum derivative ki definition (difference quotient ki limit) me f + g f+g f + g f f f g g g
Practical fayda? Har polynomial, har sin + e x + ln \sin + e^x + \ln sin + e x + ln 3 x 4 − 5 x 2 + 7 3x^4 - 5x^2 + 7 3 x 4 − 5 x 2 + 7 12 x 3 − 10 x 12x^3 - 10x 12 x 3 − 10 x 7 7 7 0 0 0
Sabse bada trap: product me yeh trick mat lagao! ( f g ) ′ ≠ f ′ g ′ (fg)' \ne f'g' ( f g ) ′ = f ′ g ′ + + + − - −
