Logarithms & Differentiation for NEET — basics to advanced

📏 Logarithms — from basics

Logarithm is just a backwards way to write a power. If powers ask "what do I get when I multiply 10 by itself 3 times?", logarithms ask "how many times do I multiply 10 to get 1000?". Same question, just looked at from the other end.

1.1 The big idea

If a^x = N ⇔ log_a N = x

🎈 Plain English

"What power do I raise a to, to get N?" — that answer is the logarithm of N to base a. So log₁₀ 1000 = 3 means "10 multiplied by itself 3 times gives 1000". Logs are just questions about powers, written in a tidy short form.

The three quantities

Base (a) — the number being multiplied
Argument (N) — the result you reached
Log value (x) — how many times the base was multiplied

1.2 Two log bases you must know

Name	Symbol	Base	Where it appears
Common log	`log N` (or log₁₀)	10	pH, decibels, Richter scale, Nernst
Natural log	`ln N` (or log_e)	e ≈ 2.718	Radioactive decay, Arrhenius, growth

ln N = 2.303 × log₁₀ N

🎈 Why 2.303?

To switch between the two bases, multiply by 2.303 (which is just ln 10). Memorise this — it's the most-used conversion in chemistry kinetics.

1.3 Quick reading exercise

Expression	Means	Value
`log 100`	10 multiplied with itself ? times = 100	2
`log 1`	10 multiplied 0 times	0
`log 0.01`	10⁻²	−2
`log 0`	impossible (no power of 10 is 0)	undefined / −∞
`log of negative`	impossible (positive base only)	undefined (in real numbers)
`ln e`	e raised to power 1	1

📏 All the Log Rules — complete reference

Every log rule you will ever need for NEET / JEE, organised by what they do. Learn them in any order — they're not independent. The product rule, power rule and change-of-base are the three you'll use most.

2.1 Basic identities (start here)

Rule	Formula	Plain meaning
Log of 1	`log_a 1 = 0`	Anything to power 0 is 1, so its log is 0.
Log of base	`log_a a = 1`	Base × itself one time is the base.
Log of base power	`log_a(a^x) = x`	The base "cancels" — read off the exponent.
Inverse identity	`a^{(log_a x)} = x`	Power and log undo each other.
Equality rule	If `log_a x = log_a y`, then x = y	Logs are one-to-one (no two inputs give the same output).

2.2 The operation rules — multiplication, division, reciprocal

Rule	Formula	Plain meaning
Product	`log(xy) = log x + log y`	Multiply two numbers → ADD their logs.
Quotient	`log(x/y) = log x − log y`	Divide → SUBTRACT logs.
Reciprocal	`log(1/x) = − log x`	Special case of quotient: numerator is 1, so log 1 = 0 disappears.
Negative power	`log(x⁻ⁿ) = −n · log x`	Same as power rule with negative exponent.

🎈 Plain English

Multiplication is hard, addition is easy. Logs turn multiplications INTO additions and divisions into subtractions. Old slide-rules used this trick to do calculations before computers existed.

2.3 Power and root rules

Rule	Formula	Plain meaning
Power rule	`log(xⁿ) = n · log x`	Pull the power OUT to the front as a multiplier.
Root rule	`log(ⁿ√x) = (1/n) · log x`	Same as power rule with fractional exponent (1/n).
Power in base	`log_aⁿ(x) = (1/n) · log_a x`	Raise the BASE to a power → divide the log by n.
Power on both	`log_aⁿ(x^m) = (m/n) · log_a x`	Combination of the two above.

2.4 Change of base & base swapping

Rule	Formula	Plain meaning
Change of base	`log_b x = (log_a x) / (log_a b)`	Convert from any base to any other — divide by the log of the new base.
Swap base & argument	`log_a b = 1 / log_b a`	Reciprocals of each other when base and argument are swapped.
Three-base chain	`log_a b · log_b c = log_a c`	Chain rule for logs — middle base cancels.
Product of swaps	`log_a b · log_b a = 1`	Special case of swap rule.

2.5 Natural log ↔ Common log conversions

Rule	Formula	Plain meaning
ln to log	`ln x = 2.303 × log₁₀ x`	Most-used conversion in chemistry kinetics.
log to ln	`log₁₀ x = 0.4343 × ln x`	= ln x / 2.303 = ln x × log e.
ln of e^x	`ln(e^x) = x`	Natural log and exponential cancel.
e to the ln	`e^{(ln x)} = x`	The other direction — also cancel.
10 to the log	`10^{(log x)} = x`	Antilog of log of x is x.

2.6 Sign & domain rules

Situation	Sign / Value	Why
`x > 1`	`log₁₀ x > 0` (positive)	x is bigger than the base raised to power 0 (= 1).
`0 < x < 1`	`log₁₀ x < 0` (negative)	Need a negative power of 10 to shrink past 1.
`x = 1`	`log_a 1 = 0`	Any base to power 0 = 1.
`x = 0`	Undefined / −∞	No finite power makes a positive base equal 0.
`x < 0`	Undefined (in real numbers)	No real exponent of a positive base gives a negative.
Base `a ≤ 0` or `a = 1`	Not allowed	Base must be positive and ≠ 1.

2.7 Inequality (monotonicity) rules

If…	Then…
Base `a > 1` AND `x > y > 0`	`log_a x > log_a y` (log is increasing)
Base `0 < a < 1` AND `x > y > 0`	`log_a x < log_a y` (log is decreasing — flips!)
For ANY base, `x = y`	`log x = log y`

🎈 Plain English

When the base is bigger than 1 (like 10 or e), the log function is "increasing" — bigger input gives bigger output. When the base is between 0 and 1 (rare, but exists), the function FLIPS — bigger input gives smaller log. Standard NEET / JEE always uses base > 1, so don't sweat the second case unless explicitly asked.

2.8 Useful derived results

Rule	Formula
Log of product chain	`log(x₁ · x₂ · x₃ · …) = Σ log xᵢ`
Log of power chain	`log(x₁^a₁ · x₂^a₂) = a₁ log x₁ + a₂ log x₂`
Log of square root	`log √x = (1/2) log x`
Log of nth root of mth power	`log(x^m/n) = (m/n) log x`
Sum of consecutive integers logs	`log 1 + log 2 + log 3 + … + log n = log(n!)`

2.9 Worked tiny examples — applying each rule

📌 Product rule

log 12 = log(4 × 3) = log 4 + log 3 = 2 log 2 + log 3 = 2(0.30) + 0.48 = 1.08

📌 Quotient + Reciprocal

log 0.5 = log(1/2) = − log 2 = −0.30

📌 Power rule

log(8³) = 3 · log 8 = 3 · 3 log 2 = 9 × 0.30 = 2.71

📌 Root rule

log √1000 = (1/2) · log 1000 = (1/2) × 3 = 1.5

📌 Change of base

log₂ 32 = log 32 / log 2 = (5 log 2) / log 2 = 5 (since 2⁵ = 32)

📌 Power in base

log₉ 27 = log_3²(3³) = (3/2) · log₃ 3 = 3/2

📌 Swap rule

log₂ 8 · log₈ 2 = 3 · (1/3) = 1 (matches log_a b · log_b a = 1)

📌 Chain rule

log₂ 4 · log₄ 16 = log₂ 16 = 4

📌 ln ↔ log conversion

ln 100 = 2.303 × log 100 = 2.303 × 2 = 4.606

📌 Inverse cancellation

10^{(log 25)} = 25 directly. e^{(ln 7)} = 7 directly.

⚡ Memory aid: The three rules you will reach for in 90% of NEET problems are product, power, and ln ↔ log conversion (×2.303). Everything else is a special case of these.

📏 Log values you MUST memorise

Memorise just three numbers (log 2, log 3, log 7) and you can compute almost every log you'll meet in NEET / JEE.

Number	log₁₀ value	How to derive
1	0	10⁰ = 1
2	0.3010	memorise
3	0.4771	memorise
4	0.6020	= 2 × log 2
5	0.6990	= 1 − log 2 (since 5 = 10/2)
6	0.7781	= log 2 + log 3
7	0.8451	memorise
8	0.9030	= 3 × log 2
9	0.9542	= 2 × log 3
10	1	10¹ = 10

3.1 Natural-log values

Constant	Value
`ln 2`	0.693
`ln 3`	1.099
`ln 10`	2.303
`ln e`	1
`e`	2.718

3.2 Antilog mental estimates

"Antilog" just means "go from the log back to the number". antilog 0.30 ≈ 2 (since log 2 ≈ 0.30).

If log x ≈	then x ≈
0	1
0.30	2
0.48	3
0.70	5
0.90	8
1	10
2	100

⚡ Characteristic and mantissa. Any log has two parts. log 250 = 2.398: the 2 (characteristic) comes from "how many digits before decimal − 1" (250 has 3 digits → 2). The 0.398 (mantissa) is found from a log table for 2.5 (since 250 = 2.5 × 10²).

📏 Where logarithms appear in NEET

Every chapter you'll meet has at least one log formula. Recognise them at first sight.

Chemistry — Ionic Equilibrium

pH = −log [H⁺] | pK_a = −log K_a

🧪 Where you use it

If [H⁺] = 10⁻³, take the log: −(−3) = 3. The minus signs make small numbers turn into easy-to-read ones (pH 1 to 14).

Chemistry — Chemical Kinetics (Arrhenius)

ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)

log(k₂/k₁) = (E_a/2.303 R)(1/T₁ − 1/T₂)

🎈 Why a log appears

Reaction rate "k" doubles, triples, sometimes goes up a million times as you heat things. Big jumps like that fit neatly on a log scale — the formula stays one line tall instead of running off the page.

Chemistry — Electrochemistry (Nernst)

E = E° − (0.0591/n) log Q (at 298 K)

Chemistry — Thermodynamics

ΔG° = −2.303 RT log K ⇒ log K = −ΔG°/(2.303 RT)

Physics — Modern Physics & Decay

N = N₀ · e^−λt ⇒ ln(N/N₀) = −λt

t_½ = ln 2 / λ = 0.693/λ

Physics — Sound (Intensity Level)

β (dB) = 10 log(I/I₀)

Worked NEET — Arrhenius

NEET-styleA reaction's rate constant doubles when temperature rises from 300 K to 310 K. Activation energy is ≈ ?

Show solution

Use log(k₂/k₁) = (E_a/2.303 R) · (1/T₁ − 1/T₂).

Left side: log 2 = 0.301.

Right factor: (1/300 − 1/310) = 10/(300·310) ≈ 1.075×10⁻⁴ K⁻¹.

So E_a = (0.301 × 2.303 × 8.314)/(1.075×10⁻⁴) ≈ 53.6 kJ/mol.

Answer: ≈ 53 kJ/mol.

∂ Differentiation — from basics

Differentiation finds "how fast a quantity is changing" at every instant. The slope of a line tells you that. The slope of a curve changes from point to point — differentiation finds the slope at any single point.

4.1 Slope of a straight line — the first step

slope = rise/run = (y₂ − y₁)/(x₂ − x₁) = Δy/Δx

🎈 Plain English

Take two points on a line. Find how much you went UP (rise) and how much you went RIGHT (run). Divide. That number tells you how steep the line is. A bigger slope = a steeper line.

4.2 What about a curve?

A curve isn't straight, so it has a different slope at every point. We zoom in — closer and closer — until the tiny bit of curve we see looks almost straight. Compute the slope there. That's the derivative at that point.

dy/dx = lim_{Δx → 0} (y(x+Δx) − y(x))/Δx

🎈 Plain English

Imagine running a tiny ruler along the curve. As the ruler shrinks down to almost a point, the steepness of the ruler is the instantaneous slope. We don't actually do the limit by hand each time — we use ready-made rules below.

4.3 Notation

dy/dx — the derivative of y with respect to x (Leibniz)
f'(x) — same, "f-prime of x" (Lagrange)
ẏ — derivative with respect to time (physics, Newton)
All three mean the same thing in a NEET question.

∂ The Five Rules of Differentiation

Rule	Formula	Plain idea
Constant	`d/dx (c) = 0`	Numbers don't change → slope is 0.
Power	`d/dx (xⁿ) = n · xⁿ⁻¹`	Pull the power down, drop it by 1.
Sum	`d/dx (u + v) = du/dx + dv/dx`	Differentiate each piece separately.
Product	`d/dx (uv) = u'v + uv'`	"First × derivative of second + second × derivative of first."
Quotient	`d/dx (u/v) = (u'v − uv')/v²`	Mind the minus sign and the v² in the denominator.
Chain	`d/dx (f(g(x))) = f'(g(x)) · g'(x)`	Differentiate the outer, then multiply by the derivative of the inner.

🎈 Power rule — easy memory trick

For xⁿ: "drop the power to the front, then take one off the power". So x³ → 3x², x → 1, x⁵ → 5x⁴. The constant disappears because it's like multiplying by 1.

5.1 Tiny worked steps

📌 d/dx of 5x³ + 2x − 7

Power rule on each term: d/dx(5x³) = 15x², d/dx(2x) = 2, d/dx(7) = 0.
Add them up: = 15x² + 2.

📌 d/dx of (x² + 1)(x − 3) — product rule

u = x² + 1 → u' = 2x. v = x − 3 → v' = 1.
(u'v + uv') = 2x(x − 3) + (x² + 1)(1)
= 2x² − 6x + x² + 1 = 3x² − 6x + 1

📌 d/dx of sin(3x²) — chain rule

Outer = sin(stuff). Its derivative is cos(stuff).
Inner = 3x². Its derivative is 6x.
Multiply: d/dx = cos(3x²) · 6x = 6x cos(3x²).

∂ Standard Derivatives — memorise this table

f(x)	f′(x)	f(x)	f′(x)
`c`	0	`xⁿ`	`n xⁿ⁻¹`
`e^x`	`e^x`	`a^x`	`a^x ln a`
`ln x`	`1/x`	`log_a x`	`1/(x ln a)`
`sin x`	`cos x`	`cos x`	`−sin x`
`tan x`	`sec²x`	`cot x`	`−cosec²x`
`sec x`	`sec x · tan x`	`cosec x`	`−cosec x · cot x`
`sin⁻¹ x`	`1/√(1 − x²)`	`cos⁻¹ x`	`−1/√(1 − x²)`
`tan⁻¹ x`	`1/(1 + x²)`	`√x`	`1/(2√x)`
`1/x`	`−1/x²`	`\|x\|`	`x/\|x\|` (sign)

🎈 Plain English

Some functions are "their own derivative" (like e^x). Sine becomes cosine; cosine becomes minus-sine; tangent becomes "sec-squared". Memorise the table — every NEET / JEE question stitches these together.

6.1 Higher-order derivatives

Differentiate twice → you get the second derivative. Velocity is the first derivative of position. Acceleration is the second.

v = dx/dt a = dv/dt = d²x/dt²

⚛️ Differentiation in Physics

Almost every Physics formula with "rate" or "instantaneous" requires a derivative.

7.1 Kinematics

v = dx/dt a = dv/dt

⚛️ Example

If x(t) = 5t² − 3t + 2, then v = dx/dt = 10t − 3 and a = d²x/dt² = 10. Constant acceleration → uniformly accelerated motion.

7.2 Electromagnetism

EMF ε = − dΦ/dt (Faraday's Law)

I = dQ/dt (current = rate of charge flow)

V_L = L · dI/dt (voltage across inductor)

7.3 Oscillations & Waves

If x(t) = A sin(ωt), then v = Aω cos(ωt), a = −Aω² sin(ωt) = −ω² x

🎈 Why "minus omega² x"

In SHM, the acceleration always points back to the centre, so it has a negative sign. The formula a = −ω²x drops out the moment you differentiate sine twice.

7.4 Maxima and minima (turning points)

Set df/dx = 0 and solve. That gives the locations where the function reaches a peak or valley. To tell which: check the second derivative.

If f''(x) < 0 → maximum (curve bends down)
If f''(x) > 0 → minimum (curve bends up)

⚛️ NEET classic

Maximum range of a projectile launched at angle θ with speed u: R = (u²/g) sin(2θ). Take dR/dθ = 0 → cos(2θ) = 0 → θ = 45°. That's why 45° gives the longest throw.

🧪 Differentiation in Chemistry

8.1 Chemical Kinetics — rate definition

Rate = −d[reactant]/dt = +d[product]/dt

🧪 Plain English

"Rate" of a reaction is "how fast concentration is changing". Reactants disappear (minus sign), products appear (plus sign). The minus and plus keep the rate a positive number.

8.2 Differentiating the integrated rate laws

Order	Integrated form	Differentiate to get rate
Zero	`[A] = [A]₀ − kt`	`d[A]/dt = −k`
First	`ln[A] = ln[A]₀ − kt`	`d[A]/dt = −k[A]`
Second	`1/[A] = 1/[A]₀ + kt`	`d[A]/dt = −k[A]²`

8.3 Thermodynamics

(∂G/∂T)_p = −S (∂G/∂p)_T = V

📝 Worked NEET-level Examples

Logarithm · ChemistryIf pH of a solution is 3.5, then [H⁺] ≈ ?

Show solution

[H⁺] = 10⁻³·⁵ = 10⁻⁴ × 10⁰·⁵ = 10⁻⁴ × √10 = 10⁻⁴ × 3.16 = 3.16 × 10⁻⁴ M.

Plain English: pH 3.5 means somewhere between pH 3 (10⁻³) and pH 4 (10⁻⁴). The half-step inside makes it about 3× weaker than pH 3.

Logarithm · ChemistryFor a first-order reaction, half-life is 1386 s. Find rate constant k.

Show solution

k = 0.693 / t_½ = 0.693 / 1386 = 5 × 10⁻⁴ s⁻¹

Differentiation · PhysicsPosition of a particle: x(t) = 2t³ − 6t² + 9. Find velocity and acceleration at t = 2 s.

Show solution

v = dx/dt = 6t² − 12t → at t = 2: v = 24 − 24 = 0 m/s.

a = dv/dt = 12t − 12 → at t = 2: a = 12 m/s².

Plain English: at t = 2, the particle momentarily stops moving (v = 0) but is still being accelerated forward.

Differentiation · PhysicsFor SHM x = 5 sin(2t), find max velocity and max acceleration.

Show solution

A = 5, ω = 2.

Max velocity = Aω = 5 × 2 = 10 units/s

Max acceleration = Aω² = 5 × 4 = 20 units/s²

Differentiation · OptimumA wire of length 12 m is bent into a rectangle. What dimensions give maximum area?

Show solution

2(L + B) = 12 → L + B = 6 → B = 6 − L

Area A = L · B = L(6 − L) = 6L − L²

dA/dL = 6 − 2L = 0 → L = 3, B = 3 (square!)

Max area = 9 m². Plain English: for a fixed perimeter, a square always beats any other rectangle. Differentiation just proves it.

Both · ChemistryA first-order reaction is 25% complete in 60 s. When will it be 75% complete?

Show solution

25% complete → 75% left → fraction = 0.75 = (1/2)ⁿ; not whole half-lives. Use the integrated form:

k = (2.303/t) log([A₀]/[A]) = (2.303/60) log(100/75) = (2.303/60)(0.1249) = 4.79 × 10⁻³ s⁻¹

For 75% complete → [A]/[A₀] = 0.25 → log 4 = 0.602.

t = (2.303 / 4.79×10⁻³) × 0.602 ≈ 290 s

Logarithm · PhysicsA radioactive nucleus decays from 10⁶ to 1.25 × 10⁵ atoms in 30 minutes. Find half-life.

Show solution

Fraction left = 1.25 × 10⁵ / 10⁶ = 1/8 = (1/2)³ → 3 half-lives.

3 t_½ = 30 min → t_½ = 10 min.

Plain English: when the fraction is a clean power of ½, skip the log — just count.

Differentiation · OpticsLens equation 1/v − 1/u = 1/f, with f constant. Find dv/du.

Show solution

Differentiate both sides w.r.t. u: −v⁻²·(dv/du) + u⁻² = 0

So dv/du = v²/u² = (magnification)².

Plain English: if you move the object a tiny bit, the image moves by a factor m² times. That's why moving an object near a powerful lens makes the image jump.

∂ Logarithmic Differentiation — advanced

When the function looks like (something)^something — variable in BOTH the base and the exponent — neither the power rule nor the exponential rule works alone. The fix: take the log of both sides first, then differentiate. This trick turns ugly powers into easy products. Classic JEE-Advanced / advanced-physics problem type, also tested for linear charge density λ = dq/dℓ along a wire.

🎈 Plain English

You can't beat a tower of letters with the usual rules. So you "translate" the tower with a logarithm: powers come down as multipliers, and the whole expression turns into a sum that's safe to differentiate term by term.

9.1 The recipe (three steps)

Take natural log of both sides: ln q = (exponent) · ln(base).
Differentiate both sides w.r.t. the variable (use product rule + chain rule on the right).
Multiply through by q to isolate dq/dℓ.

9.2 Three NEET / JEE-style problems on linear charge density (λ = dq/dℓ)

JEE Advanced · Logarithmic Differentiation(1) Charge on a wire is q = (sin ℓ)^{sin ℓ}. Find the linear charge density at ℓ = π/6.

Show full solution

Step 1 — take ln of both sides:

ln q = sin ℓ · ln(sin ℓ)

Step 2 — differentiate (product rule on the right):

(1/q) · dq/dℓ = cos ℓ · ln(sin ℓ) + sin ℓ · (1/sin ℓ) · cos ℓ

= cos ℓ · ln(sin ℓ) + cos ℓ = cos ℓ · [ln(sin ℓ) + 1]

Step 3 — multiply by q:

λ = dq/dℓ = (sin ℓ)^{sin ℓ} · cos ℓ · [1 + ln(sin ℓ)]

Evaluate at ℓ = π/6:

sin(π/6) = 1/2, cos(π/6) = √3/2, (sin ℓ)^{sin ℓ} = (1/2)^1/2 = 1/√2, ln(1/2) = −ln 2 ≈ −0.693.

λ = (1/√2) · (√3/2) · (1 − ln 2) = (√3 / 2√2) · (1 − ln 2)

Numerically: ≈ 0.612 × 0.307 ≈ 0.188 C/m.

🎈 Plain English: Charge along the wire isn't uniform — it ripples up and down with sin ℓ. To get how fast it changes at a particular point, log-and-differentiate is the only way. At π/6 the wire has about 0.19 coulombs per metre at that exact spot.

JEE Advanced · Logarithmic Differentiation(2) Charge on a wire is q = (log_e ℓ)^ℓ. Find the linear charge density.

Show full solution

Step 1 — take ln of both sides:

ln q = ℓ · ln(ln ℓ)

Step 2 — differentiate (product rule + chain rule):

(1/q) · dq/dℓ = 1 · ln(ln ℓ) + ℓ · (1/ln ℓ) · (1/ℓ)

= ln(ln ℓ) + 1/(ln ℓ)

Step 3 — multiply by q:

λ = dq/dℓ = (ln ℓ)^ℓ · [ln(ln ℓ) + 1/(ln ℓ)]

🎈 Plain English: Here the variable ℓ sits in BOTH the base and the exponent — that's why the logarithm trick is forced on us. The answer has two pieces: a "drop-the-power" part and a "logarithm of a logarithm" part.

JEE Advanced · Logarithmic Differentiation(3) Charge on a wire is q = ℓ^{ℓ^ℓ} (a double power tower). Find the linear charge density at ℓ = 2 m.

Show full solution

Step 1 — take ln of both sides:

ln q = ℓ^ℓ · ln ℓ

Step 2 — differentiate the right side. Need d(ℓ^ℓ)/dℓ first. Let u = ℓ^ℓ:

ln u = ℓ · ln ℓ ⇒ (1/u) · du/dℓ = ln ℓ + 1 ⇒ du/dℓ = ℓ^ℓ · (1 + ln ℓ)

Now apply product rule to ℓ^ℓ · ln ℓ:

(1/q) · dq/dℓ = (du/dℓ) · ln ℓ + u · (1/ℓ) = ℓ^ℓ(1 + ln ℓ) · ln ℓ + ℓ^ℓ/ℓ

= ℓ^ℓ · [ln ℓ · (1 + ln ℓ) + 1/ℓ] = ℓ^ℓ · [(ln ℓ)² + ln ℓ + 1/ℓ]

Step 3 — multiply by q:

λ = dq/dℓ = ℓ^{ℓ^ℓ} · ℓ^ℓ · [(ln ℓ)² + ln ℓ + 1/ℓ]

Evaluate at ℓ = 2:

ℓ^ℓ = 2² = 4 | ℓ^{ℓ^ℓ} = 2⁴ = 16 | ln 2 = 0.693

Bracket: (0.693)² + 0.693 + 0.5 = 0.480 + 0.693 + 0.500 = 1.673

λ = 16 × 4 × 1.673 ≈ 107.1 C/m

🎈 Plain English: A power tower grows scarily fast — 2² = 4 but 2⁴ = 16. Its rate of growth (the charge density) is even faster: about 107 C per metre at ℓ = 2. The two-step trick of "log, log again" tames the tower one storey at a time.

9.3 General pattern to remember

If q = u(ℓ)^v(ℓ), then dq/dℓ = q · [v'(ℓ) · ln u + v(ℓ) · u'(ℓ)/u(ℓ)]

🎈 Plain English

Whenever you see "(something)^something", quote the magic line: "log first, differentiate second, multiply by q last". Three steps. Always works.

🔑 One-Page Summary (cover this with your hand & recite)

Logarithms

Fact	Value
`log 2`	0.30
`log 3`	0.48
`log 7`	0.85
`ln 2`	0.693
`ln 10`	2.303
log of product	sum of logs
log of power	multiply log by exponent
change of base	divide by log of base
pH	−log[H⁺]
Half-life (1st order)	0.693/k
Arrhenius	ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)
Nernst	E = E° − (0.0591/n) log Q

Differentiation

Function	Derivative
`c`	0
`xⁿ`	`n xⁿ⁻¹`
`e^x`	`e^x`
`ln x`	`1/x`
`sin x`	`cos x`
`cos x`	`−sin x`
`tan x`	`sec²x`
`1/x`	`−1/x²`
Product	`u'v + uv'`
Quotient	`(u'v − uv')/v²`
Chain	`f'(g(x)) · g'(x)`
v (velocity)	dx/dt
a (acceleration)	dv/dt = d²x/dt²
Faraday EMF	−dΦ/dt
Reaction rate	−d[A]/dt
Max / min	df/dx = 0; check f''

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