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The Derivative & Its Rules · Topic 06 of 23

The Derivative: Definition and Interpretation

The derivative f′(a)f'(a) is one number: the limit of the secant slopes f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h} as h→0h\to 0. Geometrically it is the slope of the tangent line at x=ax=a; physically it is the instantaneous rate of change of ff there, in units of ff per unit of xx.

4 min readCore — on every final6 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    Secant slopes become the tangent slope

    f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h} is the slope of the secant through (a,f(a))(a,f(a)) and (a+h,f(a+h))(a+h,f(a+h)). Let h→0h\to0 and the secants settle onto the tangent — its slope is f′(a)f'(a).

  2. 2

    Two forms, one limit

    lim⁡h→0f(a+h)−f(a)h\lim_{h\to0}\frac{f(a+h)-f(a)}{h} and lim⁡x→af(x)−f(a)x−a\lim_{x\to a}\frac{f(x)-f(a)}{x-a} are the same limit (put x=a+hx=a+h). Use whichever makes the algebra cancel.

  3. 3

    Cancel hh before you let h→0h\to0

    Substituting h=0h=0 first gives 00\frac00. Expand, factor out hh (or use a conjugate, or combine fractions), cancel, and only then take the limit.

  4. 4

    Differentiable ⇒ continuous, never the reverse

    A corner (∣x∣|x| at 00), a cusp (x2/3x^{2/3}), a vertical tangent (x1/3x^{1/3}) or any discontinuity kills f′(a)f'(a). Continuity alone never guarantees a derivative.

  5. 5

    Read the units and the sign

    If VV is in litres and tt in minutes, V′(10)=−3.5V'(10)=-3.5 means the tank is losing 3.53.5 litres per minute at t=10t=10. Sign gives direction, size gives steepness.

-0.50.51.01.52.02.53.02468xysecanttangent, slope f′(a) = 2(a, f(a))(a+h, f(a+h))y = x²
As h→0h\to0 the second point slides toward (a,f(a))(a,f(a)) and the secant (dashed) rotates onto the tangent (amber). The tangent's slope is f′(a)f'(a) — here f(x)=x2f(x)=x^2, a=1a=1, f′(1)=2f'(1)=2.
-2.0-1.5-1.0-0.50.51.01.52.0-0.50.51.01.52.0xyslope −1slope +1corner: f′(0) does not existy = |x|
Continuous at 00 but not differentiable there: the one-sided derivatives are f−′(0)=−1f'_-(0)=-1 and f+′(0)=1f'_+(0)=1. Unequal one-sided slopes mean no tangent line.
02

Formulas

What to have memorised

  • Derivative at a point (hh-form)

    f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}

    The limit must be two-sided and finite.

  • Derivative at a point (xx-form)

    f′(a)=lim⁡x→af(x)−f(a)x−af'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}

    Same limit; handy when f(x)−f(a)f(x)-f(a) factors.

  • Tangent line at x=ax=a

    y=f(a)+f′(a) (x−a)y=f(a)+f'(a)\,(x-a)

    The slope is the number f′(a)f'(a), not the function f′(x)f'(x).

  • Normal line

    y=f(a)−1f′(a) (x−a)y=f(a)-\frac{1}{f'(a)}\,(x-a)

    Needs f′(a)≠0f'(a)\neq 0; if f′(a)=0f'(a)=0 the normal is the vertical line x=ax=a.

  • Linear estimate

    f(a+Δx)≈f(a)+f′(a) Δxf(a+\Delta x)\approx f(a)+f'(a)\,\Delta x

    Good for small Δx\Delta x — the tangent stands in for the curve.

  • One-sided derivatives

    f±′(a)=lim⁡h→0±f(a+h)−f(a)hf'_{\pm}(a)=\lim_{h\to 0^{\pm}}\frac{f(a+h)-f(a)}{h}

    f′(a)f'(a) exists exactly when both exist, are finite, and agree.

03

Differentiate from the definition

The steps, in order
  1. 1

    Write f(x+h)f(x+h) by replacing every xx with x+hx+h, then expand fully.

  2. 2

    Form f(x+h)−f(x)f(x+h)-f(x) and simplify — every term without an hh must cancel.

  3. 3

    Get hh out of the denominator: factor hh from the numerator, or multiply by a conjugate for roots, or combine fractions.

  4. 4

    Only now let h→0h\to0. Keep writing lim⁡h→0\lim_{h\to0} on every line until it is gone.

  5. 5

    For a tangent line, evaluate: f(a)f(a) is the point, f′(a)f'(a) is the slope, then y=f(a)+f′(a)(x−a)y=f(a)+f'(a)(x-a).

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    f(x+h)=f(x)+hf(x+h)=f(x)+h

    ✓ Right

    Replace every xx: for f(x)=x2−4xf(x)=x^2-4x, f(x+h)=(x+h)2−4(x+h)f(x+h)=(x+h)^2-4(x+h).

    Why: hh is added to the input, not to the output.

  • ✗ Wrong

    Putting h=0h=0 into f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h} and writing 00\frac00

    ✓ Right

    Simplify until the hh in the denominator cancels, then take the limit.

    Why: The quotient is undefined at h=0h=0; the limit is about nearby hh.

  • ✗ Wrong

    x+h=x+h\sqrt{x+h}=\sqrt{x}+\sqrt{h}

    ✓ Right

    Multiply top and bottom by the conjugate x+h+x\sqrt{x+h}+\sqrt{x}.

  • ✗ Wrong

    Tangent line y=f′(x) (x−a)y=f'(x)\,(x-a), or y=f′(a) xy=f'(a)\,x

    ✓ Right

    y=f(a)+f′(a)(x−a)y=f(a)+f'(a)(x-a): a number for the slope, through the actual point.

    Why: The line must pass through (a,f(a))(a,f(a)) with slope the number f′(a)f'(a).

  • ✗ Wrong

    "ff is continuous at aa, so f′(a)f'(a) exists."

    ✓ Right

    Differentiable ⇒ continuous only. ∣x∣|x| is continuous at 00 with no derivative there.

  • ✗ Wrong

    V′(10)=−3.5V'(10)=-3.5 litres

    ✓ Right

    V′(10)=−3.5V'(10)=-3.5 litres per minute, and the volume is decreasing.

    Why: A derivative's units are (units of ff) per (unit of xx).

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    Use the limit definition to find f′(x)f'(x) for f(x)=4−3x2f(x)=4-3x^2, then give an equation of the tangent line at x=2x=2.

    Hint

    Expand f(x+h)f(x+h), subtract f(x)f(x), and look for the factor of hh that must cancel.

    Show answer

    Answer

    f′(x)=−6xf'(x)=-6x; tangent line y=−12x+16y=-12x+16.

    Steps

    f(x+h)=4−3(x+h)2=4−3x2−6xh−3h2f(x+h)=4-3(x+h)^2=4-3x^2-6xh-3h^2, so f(x+h)−f(x)=−6xh−3h2=h(−6x−3h)f(x+h)-f(x)=-6xh-3h^2=h(-6x-3h).

    f′(x)=lim⁡h→0h(−6x−3h)h=lim⁡h→0(−6x−3h)=−6xf'(x)=\lim_{h\to0}\frac{h(-6x-3h)}{h}=\lim_{h\to0}(-6x-3h)=-6x

    At x=2x=2: f(2)=−8f(2)=-8 and f′(2)=−12f'(2)=-12, so y=−8−12(x−2)=−12x+16y=-8-12(x-2)=-12x+16. Check: x=2x=2 gives −8=f(2)-8=f(2). ✓

  2. Q2easy

    Each limit is a derivative f′(a)f'(a) in disguise. Identify ff and aa, then evaluate.

    (a) lim⁡h→0(3+h)4−81h\displaystyle\lim_{h\to0}\frac{(3+h)^4-81}{h} (b) lim⁡x→2x3−8x−2\displaystyle\lim_{x\to2}\frac{x^3-8}{x-2}

    Hint

    Match each to f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h} or f(x)−f(a)x−a\frac{f(x)-f(a)}{x-a} and ask what number is playing aa.

    Show answer

    Answer

    (a) f(x)=x4f(x)=x^4, a=3a=3, value 108108. (b) f(x)=x3f(x)=x^3, a=2a=2, value 1212.

    Steps

    (a) (3+h)4−81=108h+54h2+12h3+h4(3+h)^4-81=108h+54h^2+12h^3+h^4; divide by hh and let h→0h\to0 to get 108108. Check with the power rule: f′(x)=4x3f'(x)=4x^3, 4⋅27=1084\cdot 27=108.

    (b) x3−8=(x−2)(x2+2x+4)x^3-8=(x-2)(x^2+2x+4); cancel x−2x-2 and substitute x=2x=2: 4+4+4=124+4+4=12. Check: 3x23x^2 at x=2x=2 is 1212.

  3. Q3medium

    Find f′(x)f'(x) for f(x)=2x+1f(x)=\sqrt{2x+1} from the definition, and state the domain of f′f'.

    Hint

    Multiply numerator and denominator by 2x+2h+1+2x+1\sqrt{2x+2h+1}+\sqrt{2x+1}.

    Show answer

    Answer

    f′(x)=12x+1f'(x)=\dfrac{1}{\sqrt{2x+1}} for x>−12x>-\tfrac12.

    Steps

    2x+2h+1−2x+1h⋅2x+2h+1+2x+12x+2h+1+2x+1=2hh(2x+2h+1+2x+1)\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}\cdot\frac{\sqrt{2x+2h+1}+\sqrt{2x+1}}{\sqrt{2x+2h+1}+\sqrt{2x+1}}=\frac{2h}{h\left(\sqrt{2x+2h+1}+\sqrt{2x+1}\right)}

    Cancel hh and let h→0h\to0: f′(x)=222x+1=12x+1f'(x)=\dfrac{2}{2\sqrt{2x+1}}=\dfrac{1}{\sqrt{2x+1}}.

    ff is defined for x≥−12x\ge-\frac12 but f′f' only for x>−12x>-\frac12: the domain shrinks at the endpoint, where the right-hand quotient blows up.

  4. Q4medium

    A tank drains. V(t)V(t) is the volume in litres tt minutes after the start, with V(10)=180V(10)=180 and V′(10)=−3.5V'(10)=-3.5.

    (a) State the units of V′(10)V'(10) and what the value means. (b) Estimate V(10.5)V(10.5) and V(12)V(12). (c) Which estimate do you trust more?

    Show answer

    Answer

    (a) Litres per minute; at t=10t=10 water is leaving at 3.53.5 L/min. (b) V(10.5)≈178.25V(10.5)\approx178.25, V(12)≈173V(12)\approx173. (c) V(10.5)V(10.5) — the smaller step.

    Steps

    (a) Units of VV per unit of tt: litres/minute. The negative sign says VV is decreasing.

    (b) Tangent line at t=10t=10: L(t)=180−3.5 (t−10)L(t)=180-3.5\,(t-10). So L(10.5)=178.25L(10.5)=178.25 and L(12)=173L(12)=173.

    (c) The tangent tracks the curve only near t=10t=10; the half-minute estimate has far less room to drift than the two-minute one.

  5. Q5medium

    Show that f(x)=∣x−1∣f(x)=|x-1| is continuous at x=1x=1 but not differentiable there.

    Show answer

    Answer

    lim⁡x→1∣x−1∣=0=f(1)\lim_{x\to1}|x-1|=0=f(1), so continuous; the one-sided derivatives are −1-1 and 11, so f′(1)f'(1) does not exist.

    Steps

    Continuity: ∣x−1∣→0|x-1|\to0 as x→1x\to1, and f(1)=0f(1)=0.

    Derivative: f(1+h)−f(1)h=∣h∣h\dfrac{f(1+h)-f(1)}{h}=\dfrac{|h|}{h}, which equals 11 for h>0h>0 and −1-1 for h<0h<0. The one-sided limits disagree, so the two-sided limit — and f′(1)f'(1) — does not exist. The graph has a corner at (1,0)(1,0).

  6. Q6hard

    Let f(x)=x2f(x)=x^2 for x≤1x\le 1 and f(x)=mx+bf(x)=mx+b for x>1x>1. Find mm and bb so that ff is differentiable at x=1x=1.

    Hint

    Two conditions: the pieces must meet, and their slopes must agree.

    Show answer

    Answer

    m=2m=2, b=−1b=-1.

    Steps

    Continuity at 11: 12=m(1)+b1^2=m(1)+b, so m+b=1m+b=1.

    Matching slopes: from the left f−′(1)=2x∣x=1=2f'_-(1)=2x\big|_{x=1}=2; from the right f+′(1)=mf'_+(1)=m. So m=2m=2, then b=−1b=-1.

    Both conditions are needed — a continuous join with mismatched slopes is a corner, not a smooth curve.

06

On the exam

How this topic is marked

  • Definition questions are marked on process: show f(x+h)f(x+h) expanded, the cancellation of hh, and lim⁡h→0\lim_{h\to0} on every line until the limit is taken.

  • When a limit looks like (stuff)(a+h)−(stuff)(a)h\frac{(\text{stuff})(a+h)-(\text{stuff})(a)}{h}, name ff and aa and evaluate it with a derivative rule in one line.

  • Finish a tangent-line question with an equation of a line, and a rate question with units and a sentence about the sign.

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