Intuition What this page is forThe parent note proved the three rules. This page drills them until no case can surprise you. We first lay out a matrix of every kind of problem these rules can throw at you — every sign, every degenerate input, every "trap" — then we work an example for each cell. If you finish this page, you have seen the whole landscape.
Everything here rests on three facts from the parent note . Let me restate them in plain words before we use a single symbol:
We also lean on the Power rule for individual pieces: d x d [ x n ] = n x n − 1
Below is every class of problem linearity can present. Each row is a "cell"; the examples that follow are tagged with the cell they cover.
#
Cell (case class)
What makes it tricky
Example
A
Plain polynomial, all positive terms
warm-up, power rule + sum
Ex 1
B
Mixed signs / subtraction of a multi-term group
the minus sign must distribute
Ex 2
C
Constant term + constant multiple
derivative of a constant is 0
Ex 1, Ex 2
D
Negative & fractional exponents (rewrite first)
roots and 1/ x
Ex 3
E
Different function types mixed (sin, e x
linearity lets them coexist
Ex 4
F
Degenerate: constant multiple c = 0 c = 1
limiting/edge values of c
Ex 5
G
Evaluate the slope at a specific point (sign of slope)
plug a number, read geometry
Ex 6
H
Real-world word problem (rates add)
translate words → linearity
Ex 7
I
Exam twist: looks like a product, isn't
must expand before differentiating
Ex 8
Worked example Example 1 — Cell A + C: plain polynomial with a constantDifferentiate y = 2 x 3 + 5 x 2 + 4 x + 9
Forecast: before reading on, guess y ′ 9
Split the sum: y ′ = d x d [ 2 x 3 ] + d x d [ 5 x 2 ] + d x d [ 4 x ] + d x d [ 9 ] Why this step? The sum rule lets us differentiate each term on its own and add — no term interacts with another.
Pull constants out and apply the power rule term by term:
y ′ = 2 ( 3 x 2 ) + 5 ( 2 x ) + 4 ( 1 ) + 0 = 6 x 2 + 10 x + 4. Why this step? The constant multiple rule lets 2 5 x n 9 0
Verify: the constant 9
Worked example Example 2 — Cell B: subtraction of a whole groupDifferentiate y = x 4 − ( 3 x 2 − 8 x + 1 )
Forecast: what sign does the 8 x
Distribute the minus first: y = x 4 − 3 x 2 + 8 x − 1 Why this step? A leading minus is really ( − 1 ) × whole bracket. If you only negate the first term you break the difference rule , which says negate everything in g
Differentiate term by term:
y ′ = 4 x 3 − 6 x + 8 − 0 = 4 x 3 − 6 x + 8. Why this step? Sum/difference splits the terms; power rule on each; the constant − 1 0
Verify: the + 8 x + 8 not − 8 4 x 3 − 6 x − 8
Worked example Example 3 — Cell D: negative and fractional exponents (rewrite first)Differentiate y = x 2 4 + 6 x − 3 x
Forecast: you cannot use the power rule until every piece is written as c x power
Rewrite each piece as a scaled power:
y = 4 x − 2 + 6 x 1/2 − 3 1 x . Why this step? x 2 4 = 4 x − 2 x 2 x − 2 x = x 1/2 3 x = 3 1 x power rule needs an explicit exponent , so we manufacture one.
Differentiate each with the power rule, constants riding outside:
y ′ = 4 ( − 2 ) x − 3 + 6 ( 2 1 ) x − 1/2 − 3 1 ( 1 ) = − 8 x − 3 + 3 x − 1/2 − 3 1 .
Rewrite back in root/fraction form for readability:
y ′ = − x 3 8 + x 3 − 3 1 . Why this step? Negative exponents mean "in the denominator"; a − 1/2
Verify: the exponent − 2 − 3 1/2 − 1/2
Worked example Example 4 — Cell E: three different function types coexistingDifferentiate y = 5 sin x − 2 e x + 7 ln x
Forecast: do the sine, exponential, and log "interfere"? Guess before reading.
Split by linearity, constants outside:
y ′ = 5 d x d [ sin x ] − 2 d x d [ e x ] + 7 d x d [ ln x ] . Why this step? Linearity does not care what the functions are — it splits any sum of scaled pieces, whatever their type.
Insert the known derivatives ( sin x ) ′ = cos x ( e x ) ′ = e x ( ln x ) ′ = x 1 y ′ = 5 cos x − 2 e x + x 7 .
Verify: three unrelated function families sit side by side, untouched by one another — exactly the "modular calculus" the parent note promised. See Linear operators for why this always works.
Worked example Example 5 — Cell F: degenerate constantsc = 0 c = 1 y = 0 ⋅ x 5 + 3 x y = 1 ⋅ cos x − x 2
Forecast: in (a), does the x 5
Part (a): the constant multiple rule with c = 0 d x d [ 0 ⋅ x 5 ] = 0 ⋅ ( 5 x 4 ) = 0 Why this step? Scaling a function by 0 0 x 5 invisible to the derivative.
y ′ = 0 + 3 = 3.
Part (b): the constant multiple rule with c = 1 d x d [ 1 ⋅ cos x ] = 1 ⋅ ( − sin x ) = − sin x y ′ = − sin x − 2 x . Why this step? Multiplying by 1 c = 1
Verify: c = 0 c = 1
Worked example Example 6 — Cell G: read the slope's sign at a point (geometric)For y = x 3 − 3 x y ′ x = − 2 x = 0 x = 1
Forecast: at x = 0
Differentiate by linearity + power rule:
y ′ = 3 x 2 − 3. Why this step? Sum/difference splits the two terms; power rule + constant multiple handle each.
Plug in each point:
x = − 2 : y ′ = 3 ( 4 ) − 3 = 9 positive slope (rising).x = 0 : y ′ = 3 ( 0 ) − 3 = − 3 negative slope (falling).x = 1 : y ′ = 3 ( 1 ) − 3 = 0 zero slope (a flat spot / turning point).
Why this step? The derivative is itself a function; feeding it an x there .
Verify: the figure shows the curve rising on the left (slope + 9 x = 1 0 − 3
Worked example Example 7 — Cell H: a real-world word problem (rates add)A tank is filled by two pipes. Pipe A adds water at position A ( t ) = 4 t 2 B ( t ) = 3 t t total water is W ( t ) = A ( t ) + B ( t ) t = 2
Forecast: guess — is the total fill rate just A's rate plus B's rate?
"How fast" means the derivative W ′ ( t ) sum rule , the rate of the total is the sum of the individual rates:
W ′ ( t ) = A ′ ( t ) + B ′ ( t ) . Why this step? This is the physical meaning of linearity: combined rates add — exactly the two-runners picture from the parent note.
Differentiate each: A ′ ( t ) = 8 t B ′ ( t ) = 3 W ′ ( t ) = 8 t + 3
Evaluate at t = 2 W ′ ( 2 ) = 8 ( 2 ) + 3 = 19
Verify (units): 8 t 4 t 2 d / d t 3 litres/min , a valid rate. Numeric: 19
Worked example Example 8 — Cell I: exam twist thatlooks like a product
Differentiate y = ( x + 2 ) ( x − 5 )
Forecast: it's a product — do you reach for the product rule? Should you?
Expand first (linearity only handles sums, not products):
y = x 2 − 5 x + 2 x − 10 = x 2 − 3 x − 10. Why this step? The parent note warns that the derivative of a product is not the product of derivatives. But once expanded, y does apply. (Alternatively use the Product rule — you'll get the same answer.)Differentiate the sum:
y ′ = 2 x − 3.
Verify: cross-check with the Product rule ( f g ) ′ = f ′ g + f g ′ f = x + 2 , g = x − 5 ( 1 ) ( x − 5 ) + ( x + 2 ) ( 1 ) = x − 5 + x + 2 = 2 x − 3 couldn't use linearity; expanding made it legal.
Recall Self-check: which cell was hardest?Sign trap (Cell B) ::: the leading minus must distribute over every term inside the bracket.
Rewrite trap (Cell D) ::: convert roots and fractions to explicit powers before the power rule.
Product trap (Cell I) ::: expand first — linearity never splits a product.
Mnemonic The whole page in one lineSplit the sum, ride the constants, distribute the minus, and expand any product before you dare use linearity.
Sum, difference, constant multiple rules — the parent rules drilled here.Power rule — the per-term engine used in almost every example.Product rule — the sibling rule needed in Cell I.Definition of the derivative (limit of difference quotient) — where slopes come from.Limit laws (sum, scalar, product of limits) — why the sum rule is legal.Linearity of integration — the same trick, run backwards.Linear operators — the abstract reason different function types never interfere.