Mini-Lesson

Differentiation

OCR section 1.07 starts from first principles, adds the chain, product and quotient rules, then goes further: implicit and parametric differentiation, points of inflection, and connected rates of change.

Every trig derivative below assumes x is in radians. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.

Differentiation · first principles

Differentiation from first principles

The derivative is the limit of the gradient of a chord as the chord shrinks to a point.

f′(x) = lim_h→0 [ f(x + h) − f(x) ] / hOCR requires you to be able to carry this out for small positive integer powers, and for sin x and cos x

Worked example — differentiate f(x) = x³ from first principles

f(x + h) = (x + h)³ = x³ + 3x²h + 3xh² + h³.

[f(x + h) − f(x)]/h = (3x²h + 3xh² + h³)/h = 3x² + 3xh + h².

Now let h → 0: the last two terms vanish. ∴ f′(x) = 3x².

So at x = 2 the gradient is 3(2²) = 12.

Do not write h = 0 in the middle of the working — you would be dividing by zero. Divide by h first, then take the limit. That order is worth a mark.

Calculate

Your turn — first principles

1Differentiating f(x) = x³ from first principles gives f′(x) = 3x². Find the gradient of the curve at x = 2.

Hint: f′(2) = 3 × 2² = 3 × 4.

Differentiation · standard results

The standard derivatives

d/dx (xⁿ) = nxⁿ⁻¹ · d/dx (e^kx) = ke^kx · d/dx (ln x) = 1/xd/dx(sin x) = cos x · d/dx(cos x) = −sin x · d/dx(tan x) = sec²x — all in RADIANS

And the three rules that build everything else:

Chain rule: if y = f(g(x)) then dy/dx = f′(g(x)) × g′(x). Differentiate the outside, times the derivative of the inside.
Product rule: if y = uv then dy/dx = u′v + uv′.
Quotient rule: if y = u/v then dy/dx = (u′v − uv′) / v². The order matters — it is not symmetric.

The single most common error is losing the minus sign: d/dx(cos x) = −sin x, and the quotient rule has u′v minus uv′ (not the other way round).

Differentiation · chain rule

The chain rule in action

Worked example — y = (3x² + 1)⁴

Outside: ( )⁴ → 4( )³. Inside: 3x² + 1 → 6x.

dy/dx = 4(3x² + 1)³ × 6x = 24x(3x² + 1)³.

At x = 1: 3(1)² + 1 = 4, so dy/dx = 24(1)(4³) = 24 × 64 = 1536.

Worked example — quotient rule, y = (2x + 1)/(x − 3)

u = 2x + 1 (u′ = 2), v = x − 3 (v′ = 1).

dy/dx = [2(x − 3) − (2x + 1)(1)] / (x − 3)² = (2x − 6 − 2x − 1)/(x − 3)² = −7/(x − 3)².

At x = 5: −7/(2²) = −1.75. Note the gradient is negative everywhere — the curve is always decreasing.

Calculate

Your turn — chain rule

2Given y = (3x² + 1)⁴, find dy/dx when x = 1.

Hint: dy/dx = 4(3x² + 1)³ × 6x = 24x(3x² + 1)³. At x = 1 the bracket is 4, so 24 × 4³.

Calculate

Your turn — quotient rule

3Given y = (2x + 1)/(x − 3), find dy/dx when x = 5.

Hint: dy/dx = [2(x − 3) − (2x + 1)] / (x − 3)² = −7/(x − 3)². At x = 5 that is −7/4.

Quick check

Which rule?

?Which rule is the efficient first choice for differentiating y = x³ ln x?

Differentiation · stationary points

Stationary points and inflection

At a stationary point, dy/dx = 0. To classify it, use the second derivative:

d²y/dx² > 0 ⇒ minimum · d²y/dx² < 0 ⇒ maximumif d²y/dx² = 0 the test is inconclusive — check the sign of dy/dx either side instead

Worked example — y = x³ − 6x² + 9x + 2

dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3.

d²y/dx² = 6x − 12.

At x = 1: 6 − 12 = −6 < 0 ⇒ maximum. y = 1 − 6 + 9 + 2 = 6. So (1, 6).

At x = 3: 18 − 12 = +6 > 0 ⇒ minimum. y = 27 − 54 + 27 + 2 = 2. So (3, 2).

Points of inflection are where the curve changes concavity. A necessary condition is d²y/dx² = 0 and it changes sign. Here d²y/dx² = 0 at x = 2, and it does change sign, so (2, 4) is a (non-stationary) point of inflection.

Calculate

Your turn — the minimum

4The curve y = x³ − 6x² + 9x + 2 has a minimum point. Find its y-coordinate.

y =

Hint: dy/dx = 3(x − 1)(x − 3) = 0 gives x = 1 or 3. Since d²y/dx² = 6x − 12 is positive at x = 3, that is the minimum. Now find y at x = 3.

Quick check

Classifying a stationary point

?At a stationary point of y = f(x), the second derivative is −4. What is the point?

Differentiation · implicit

Implicit and parametric differentiation

Implicit: when y is not given explicitly, differentiate every term with respect to x and use the chain rule on any y term. The key line:

d/dx (y²) = 2y · dy/dx · d/dx (xy) = y + x · dy/dxthe second one needs the product rule as well

Worked example — x² + y² = 25

Differentiate: 2x + 2y(dy/dx) = 0 ⇒ dy/dx = −x/y.

At (3, 4): dy/dx = −3/4. Check: this is a circle of radius 5, and the radius to (3, 4) has gradient 4/3 — the tangent gradient −3/4 is its perpendicular. ✓

Parametric: when x and y are both given in terms of t:

dy/dx = (dy/dt) ÷ (dx/dt)the dt cancels — this is just the chain rule rearranged

Worked example — x = t², y = t³

dx/dt = 2t, dy/dt = 3t². ∴ dy/dx = 3t² / 2t = 3t/2 (for t ≠ 0).

Quick check

Implicit differentiation

?What is d/dx (y³) when differentiating implicitly with respect to x?

Differentiation · rates

Connected rates of change

When two quantities are linked, their rates are linked too — by the chain rule.

dV/dt = (dV/dr) × (dr/dt)chain together the rate you know with the relationship you have, to reach the rate you want

Worked example — an inflating spherical balloon

Air is pumped in at dV/dt = 12 cm³/s. Find dr/dt when the radius is 3 cm.

V = (4/3)πr³ ⇒ dV/dr = 4πr². At r = 3: dV/dr = 4π(9) = 36π.

dV/dt = (dV/dr)(dr/dt) ⇒ 12 = 36π × dr/dt.

dr/dt = 12/(36π) = 1/(3π) = 0.106 cm/s (3 s.f.).

Method: write down what you know (dV/dt), what you want (dr/dt), and the link (V in terms of r). The chain rule then assembles itself.

Calculate

Your turn — connected rates

5A spherical balloon is inflated at 12 cm³/s. Given V = (4/3)πr³, find dr/dt when r = 3 cm, to 3 significant figures.

cm/s

Hint: dV/dr = 4πr² = 36π at r = 3. Then dr/dt = (dV/dt) ÷ (dV/dr) = 12 ÷ 36π.

Quick check

Parametric gradient

?A curve is given parametrically by x = f(t), y = g(t). What is dy/dx?

Sort it

Which rule do you reach for?

Tap a function, then tap the rule that is the efficient first choice.

⛓️ Chain

✖️ Product

➗ Quotient

Match it

Function and derivative

Tap a function on the left, then its derivative on the right.

Function

Derivative

Recap

The big ideas to know

First principles: f′(x) = lim as h→0 of [f(x + h) − f(x)]/h — divide by h before letting h → 0

Rules: chain (outside × inside); product u′v + uv′; quotient (u′v − uv′)/v²

Trig: sin → cos, cos → −sin, tan → sec² — radians only

Stationary points: dy/dx = 0; then d²y/dx² > 0 ⇒ min, < 0 ⇒ max

Inflection: d²y/dx² = 0 and changes sign

Implicit: d/dx(y²) = 2y · dy/dx — never forget the dy/dx

Parametric: dy/dx = (dy/dt) ÷ (dx/dt)

Rates: dV/dt = (dV/dr)(dr/dt) — know, want, link

That is OCR 1.07 — and integration is about to run all of it backwards. Press Finish to see your score.

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