Additional Mathematics Classroom | Chapter 12: Differentiation of Trigonometric, Exponential and Logarithmic Functions — Advanced Chain Rule, Tangents, Rates and Optimisation | SEC G3 K341

The 2027 SEC position

G3 K341 requires derivatives of xⁿ for rational n together with sin x, cos x, tan x, e^x and ln x, as well as constant multiples, sums and differences. Product Rule, Quotient Rule and Chain Rule remain active. The same G3 calculus strand then applies differentiation to increasing/decreasing functions, stationary points, second-derivative classification, gradients, tangents, normals, connected rates and maxima/minima problems.

This means the new content is not a second calculus system. It is an expanded function library plugged into the calculus system already built.

The five core derivative facts are:

• d/dx(sin x) = cos x;
• d/dx(cos x) = −sin x;
• d/dx(tan x) = sec²x;
• d/dx(e^x) = e^x;
• d/dx(ln x) = 1/x, for x > 0.

These standard trigonometric derivatives use radian measure. That is not a notation preference. It is built into the clean derivative constants.

Official reference: SEAB · 2027 SEC G3 Additional Mathematics K341.

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What this chapter is really teaching

The main transition is from rule recall to function recognition. In Chapter 10, the learner might see (3x + 1)⁵ and ask what is inside what. Now the inner or outer function may be sine, exponential or logarithmic:

• sin(3x + 1) → trigonometric outer function + Chain Rule;
• e^x² → exponential outer function + Chain Rule;
• ln(5x − 2) → logarithmic outer function + Chain Rule;
• x²e^−x → Product Rule outside, Chain Rule inside exponential;
• sin x/(1 + cos x) → Quotient Rule plus trig derivatives;
• ln[(x² + 1)] → logarithmic derivative plus Chain Rule.

The derivative rules do not compete. They stack according to structure.

First identify the function family. Then identify the outer structure. Then follow dependencies inward.

Chapter map

1. Why radians matter in calculus
2. Derivative of sine
3. Derivative of cosine
4. Derivative of tangent
5. Derivative of e^x
6. Derivative of ln x
7. Chain Rule with trig functions
8. Chain Rule with exponential functions
9. Chain Rule with logarithmic functions
10. Product Rule with new function families
11. Quotient Rule with new function families
12. Mixed rule trees
13. Tangents and normals
14. Increasing/decreasing behaviour
15. Stationary points and optimisation
16. Second derivatives
17. Rates of change and modelling
18. Verification and graph sense
19. Error diagnosis
20. Original guided practice and worked answers

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1 · Why radians matter

In trigonometry, degrees and radians both measure angle. In calculus, radians have a special status because the clean derivative

d/dx(sin x) = cos x

holds when x is measured in radians.

Why? The derivative is built from a ratio of small changes. On the unit circle, radian measure connects arc length directly to angle: arc length s = rθ. When r = 1, the numerical angle equals arc length. That natural scaling makes the limiting ratio for sine approach exactly 1.

If an angle variable were measured numerically in degrees, an extra scale factor π/180 would appear because one degree is π/180 radians.

A useful warning

Do not mix calculator angle mode with symbolic calculus. When differentiating sin x symbolically in the standard school formula, x is interpreted in radians. When evaluating a final numerical angle, follow the units and context of the question.

This is one reason radians appeared earlier in trigonometry: they are not merely another way to write 180°. They are the angle measure naturally aligned with calculus.

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2 · Differentiating sine

The standard derivative is

d/dx(sin x) = cos x.

This identity has a strong graph interpretation. The gradient of the sine graph at each x-value is given by the cosine graph.

• At x = 0, sin x crosses upward with gradient cos0 = 1.
• At x = π/2, sin x reaches a maximum and gradient cos(π/2)=0.
• At x = π, sin x crosses downward and gradient cosπ = −1.
• At x = 3π/2, sin x reaches a minimum and gradient cos(3π/2)=0.

The derivative relationship is visible in the graph before it is used algebraically.

Worked route A

Differentiate y = 7sin x.

dy/dx = 7cos x.

Worked route B

Differentiate y = 3sin x − 4x².

dy/dx = 3cos x − 8x.

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3 · Differentiating cosine

The standard derivative is

d/dx(cos x) = −sin x.

The minus sign is essential. The cosine graph begins at a maximum when x = 0, so its tangent gradient is zero there. Immediately to the right, cosine decreases, consistent with −sin x becoming negative.

Worked route

Differentiate y = 5cos x + 2sin x.

dy/dx = −5sin x + 2cos x.

Common sign failure

A frequent mistake is d(cos x)/dx = sin x. The graph itself provides a quick check. Near x = 0, cosine decreases as x increases; a positive sine derivative would incorrectly predict increasing behaviour.

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4 · Differentiating tangent

The standard derivative is

d/dx(tan x) = sec²x.

Because sec²x is positive wherever tangent is defined, tan x is increasing on every interval between its vertical asymptotes. That graph behaviour is exactly what we observe.

Deriving tangent from quotient structure

tan x = sin x / cos x. Use Quotient Rule:

d/dx(tan x)

= [cos x(cos x) − sin x(−sin x)]/cos²x

= [cos²x + sin²x]/cos²x

= 1/cos²x

= sec²x.

The derivative formula therefore connects three previous chapters: Quotient Rule, derivatives of sine/cosine and the Pythagorean identity.

Worked route

Differentiate y = 4tan x − x.

dy/dx = 4sec²x − 1.

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5 · Differentiating e^x: the function that reproduces itself

The natural exponential function has the remarkable property

d/dx(e^x) = e^x.

The function’s instantaneous rate of change is equal to its current value. This is one reason e appears so naturally in growth and decay models.

Worked route A

Differentiate y = 6e^x.

dy/dx = 6e^x.

Worked route B · growth model meaning

If P = 500e^0.2t, then

dP/dt = 100e^0.2t = 0.2P.

The model grows at an instantaneous rate equal to 20% of its current value per time unit.

That interpretation explains the exponent coefficient: it controls proportional growth rate.

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6 · Differentiating ln x

For x > 0:

d/dx(ln x) = 1/x.

The derivative decreases as x grows. The logarithm continues to rise, but more and more slowly. That matches its graph: steep near x = 0⁺, flattening gradually for large x.

Worked route A

Differentiate y = 5ln x − 3x.

dy/dx = 5/x − 3.

Domain stays alive

ln x is defined for x > 0 in the real-number setting. Differentiating it does not expand the domain. The derivative 1/x exists for negative x algebraically, but it is not the derivative of the original real function ln x there because ln x itself was not defined there.

Inverse-function intuition

ln x is the inverse of e^x. Their graph shapes and derivative behaviours are deeply connected: exponential growth becomes steeper as values increase; logarithmic growth becomes flatter.

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7 · Chain Rule with trigonometric functions

The basic trig derivative gives the outer rule. Chain Rule multiplies by the derivative of the inside.

• d/dx[sin(g(x))] = cos(g(x))g′(x);
• d/dx[cos(g(x))] = −sin(g(x))g′(x);
• d/dx[tan(g(x))] = sec²(g(x))g′(x).

Worked route A · sin(3x + 2)

Outer derivative: cos(3x + 2).

Inner derivative: 3.

Therefore

d/dx[sin(3x + 2)] = 3cos(3x + 2).

Worked route B · cos(5 − 2x)

d/dx[cos(5 − 2x)]

= −sin(5 − 2x)(−2)

= 2sin(5 − 2x).

Two negative signs cancel. Writing both derivative layers before simplifying reduces sign errors.

Worked route C · tan(x²)

d/dx[tan(x²)] = sec²(x²)(2x).

So dy/dx = 2x sec²(x²).

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8 · Chain Rule with exponential functions

For y = e^g(x):

dy/dx = e^g(x)g′(x).

The exponential outer function remains unchanged; the inner derivative appears as a multiplier.

Worked route A · e^4x−1

d/dx(e^4x−1) = 4e^4x−1.

Worked route B · e^x²

d/dx(e^x²) = 2xe^x².

Worked route C · e^−3x

d/dx(e^−3x) = −3e^−3x.

The negative coefficient matches exponential decay: the function is positive but decreasing.

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9 · Chain Rule with logarithmic functions

For y = ln(g(x)), where g(x) > 0:

dy/dx = g′(x)/g(x).

This compact form comes directly from Chain Rule:

outer derivative = 1/g(x); multiply by inner derivative g′(x).

Worked route A · ln(3x + 5)

dy/dx = 3/(3x + 5).

Worked route B · ln(x² + 1)

dy/dx = 2x/(x² + 1).

Here x² + 1 is positive for all real x, so the logarithm is defined for every real x.

Worked route C · ln(5 − x)

dy/dx = −1/(5 − x), with original domain x < 5.

Domain before derivative

For ln(x² − 4), the real domain is x < −2 or x > 2. The formal derivative 2x/(x² − 4) exists algebraically on other points except ±2, but it is not the derivative of the real logarithmic function where the original log is undefined.

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10 · Product Rule with the expanded function library

Product Rule remains

(uv)′ = u′v + uv′.

The only change is that u and v may now be trig, exponential or logarithmic functions.

Worked route A · x sin x

Let u = x and v = sin x.

u′ = 1, v′ = cos x.

dy/dx = sin x + xcos x.

Worked route B · xe^x

dy/dx = e^x + xe^x = e^x(1 + x).

The factorised form immediately exposes a stationary candidate x = −1 because e^x is never zero.

Worked route C · x²ln x

For x > 0:

dy/dx = 2x ln x + x²(1/x)

= 2xln x + x = x(2ln x + 1).

Worked route D · x²e^−x

Product Rule outside, Chain Rule inside exponential:

y′ = 2xe^−x + x²(−e^−x)

= e^−x(2x − x²) = xe^−x(2 − x).

The factorisation is ideal for stationary-point analysis.

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11 · Quotient Rule with trig, exponential and logarithmic functions

Quotient Rule remains

(u/v)′ = (vu′ − uv′)/v².

Worked route A · sin x / x

For x ≠ 0:

y′ = [x cos x − sin x]/x².

So dy/dx = (xcos x − sin x)/x².

Worked route B · e^x/x

y′ = [x e^x − e^x]/x²

= e^x(x − 1)/x².

Worked route C · ln x / x

For x > 0:

y′ = [x(1/x) − ln x]/x²

= (1 − ln x)/x².

When rewriting is cleaner

e^x/x² can be written x⁻²e^x and differentiated with Product Rule. Both routes are valid. Choose the one whose algebra you can control most clearly.

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12 · Mixed rule trees

The hardest-looking derivative questions become manageable when the outer structure is identified before any derivative is written.

Example A · x² sin(3x)

Outer structure: Product Rule.

Inside trig factor: Chain Rule.

y′ = 2x sin(3x) + x²·3cos(3x).

y′ = 2xsin(3x) + 3x²cos(3x).

Example B · e^x²ln x

Outer structure: Product Rule.

First factor needs Chain Rule; second factor is ln x.

y′ = 2xe^x²ln x + e^x²(1/x).

Factor:

y′ = e^x²(2xln x + 1/x), for x > 0.

Example C · ln(sin x)

Outer logarithm, inner sine:

y′ = cos x / sin x

= cot x, on intervals where sin x > 0 for the real logarithm.

The compact cotangent result is useful, but the domain came from the original ln(sin x), not from cot x alone.

Example D · e^{sin x}

Outer exponential, inner sine:

y′ = e^{sin x}cos x.

Example E · sin(e^x)

Outer sine, inner exponential:

y′ = cos(e^x)e^x.

Order matters. e^{sin x} and sin(e^x) are different compositions and therefore have different derivatives.

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13 · Tangents and normals to the new curves

The tangent/normal architecture from Chapter 10 does not change:

1. find the point from the original function;
2. find the tangent gradient from the derivative;
3. use point–gradient form;
4. for the normal, use the perpendicular gradient.

Worked tangent · y = e^x at x = 0

Point: (0,1).

Derivative y′ = e^x, so gradient at x = 0 is 1.

Tangent:

y − 1 = 1(x − 0).

y = x + 1.

Worked normal · y = ln x at x = 1

Point: (1,0).

Derivative y′ = 1/x, so tangent gradient at x = 1 is 1.

Normal gradient = −1.

Normal:

y = −(x − 1).

y = −x + 1.

Worked tangent · y = sin x at x = π/3

Point: (π/3, √3/2).

Derivative = cos x, so gradient = 1/2.

Tangent:

y − √3/2 = (1/2)(x − π/3).

Keep exact values exact unless approximation is requested.

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14 · Increasing and decreasing behaviour

The sign logic remains unchanged: f′ > 0 means increasing, f′ < 0 means decreasing. The new derivative families often make sign analysis more interesting.

Example A · e^x is always increasing

f(x) = e^x.

f′(x) = e^x > 0 for all real x.

Therefore e^x is increasing everywhere.

Example B · ln x is always increasing on its domain

f′(x)=1/x > 0 for x > 0.

Therefore ln x is increasing on (0,∞).

Example C · tangent increases between asymptotes

d(tan x)/dx = sec²x > 0 wherever tangent is defined. Hence tan x is increasing on each continuous interval between vertical asymptotes.

Do not say “increasing for all real x” without acknowledging the discontinuities where tangent is undefined.

Example D · xe^−x

f′(x)=e^−x(1−x).

Since e^−x > 0, the derivative sign is controlled entirely by 1−x:

• x < 1 → increasing;
• x > 1 → decreasing.

Therefore x = 1 is a local—and over suitable domains, global—maximum.

The positive exponential factor can be ignored for sign-switch purposes because it never vanishes or changes sign.

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15 · Stationary points with exponential and logarithmic functions

The method remains:

1. differentiate;
2. solve f′(x)=0;
3. find the point coordinate;
4. classify;
5. check the domain.

Worked stationary point · xe^−x

f(x)=xe^−x.

f′(x)=e^{−x}(1−x).}

Because e^−x ≠ 0, stationary when 1−x=0:

x=1.

f(1)=1/e.

Derivative changes positive→negative, so (1,1/e) is a maximum.

Worked stationary point · ln x − x/2

Domain x > 0.

f′(x)=1/x − 1/2.

Set zero:

1/x=1/2 → x=2.

f″(x)=−1/x², so f″(2)<0.

Thus x=2 gives a maximum, with value ln2−1.

A logarithmic stationary equation may need exponential reversal

If f′(x) contains ln x and solving f′=0 gives ln x = 3, then x = e³. The equation-solving skills from the Exponential and Logarithmic Functions chapter remain active inside calculus.

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16 · Stationary points with trigonometric functions

Trigonometric stationary-point questions combine calculus with interval equation solving.

Worked route · y = sin x + cos x

Consider 0 ≤ x ≤ 2π.

y′ = cos x − sin x.

Stationary when

cos x = sin x.

For points where cos x ≠ 0, tan x = 1.

In 0 ≤ x ≤ 2π:

x = π/4, 5π/4.

y(π/4)=√2.

y(5π/4)=−√2.

Second derivative:

y″=−sin x−cos x.

At π/4, y″=−√2<0 → maximum.

At 5π/4, y″=√2>0 → minimum.

This one problem uses trig derivatives, trig equations, exact values and the second derivative test.

Interval discipline remains essential

The derivative equation may have infinitely many trigonometric solutions. The question’s interval determines which stationary points belong to the problem.

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17 · Second derivatives in the expanded function library

Second derivatives require differentiating the first derivative with the same rules.

Examples

• y = sin x → y′ = cos x → y″ = −sin x.
• y = cos x → y′ = −sin x → y″ = −cos x.
• y = e^x → y′ = e^x → y″ = e^x.
• y = ln x → y′ = 1/x → y″ = −1/x².

The equation y″ = −y for sine and cosine is a powerful structural fact: the second derivative points opposite to the displacement. That relationship later becomes central in oscillation and differential equations.

Worked mixed second derivative

For y = xe^x:

y′=e^x(1+x).

Differentiate again:

y″=e^x(1+x)+e^x

= e^x(x+2).

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18 · Rates of change in growth and oscillation models

Once a function is used as a model, its derivative gives the modelled instantaneous rate.

Exponential growth model

P(t)=1200e^0.05t.

P′(t)=60e^0.05t=0.05P(t).

At every time, the instantaneous growth rate is 5% of the current modelled population per time unit.

Exponential decay model

M(t)=80e^−0.3t.

M′(t)=−24e^−0.3t=−0.3M(t).

The negative sign records decay; the magnitude is 30% of the current amount per time unit.

Oscillation model

s(t)=4sin(2t).

Velocity:

v(t)=s′(t)=8cos(2t).

Acceleration:

a(t)=v′(t)=−16sin(2t)=−4s(t).

The acceleration is proportional to displacement and points in the opposite direction. Even though differential-equation theory is beyond the immediate chapter, the derivative relationship reveals the dynamics.

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19 · Optimisation with exponential functions

Functions such as x e^−x or x²e^−x combine polynomial growth with exponential decay. Their products often rise initially and fall later, producing a natural maximum.

Worked optimisation · x²e^−x

For x ≥ 0, find the maximum of

f(x)=x²e^−x.

Derivative:

f′(x)=xe^−x(2−x).

Since e^−x>0, stationary candidates:

x=0 or x=2.

On x≥0, x=0 is the boundary and f(0)=0.

At x=2:

f(2)=4e⁻²=4/e².

For 0<x<2, f′>0; for x>2, f′<0.

Therefore maximum = 4/e² at x=2.

The factorisation makes the optimisation almost entirely a sign problem.

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20 · Optimisation with logarithms

Logarithmic functions often appear in objectives involving diminishing returns or in expressions created by algebraic models.

Worked route · ln x / x

For x>0, consider

f(x)=ln x / x.

Derivative:

f′(x)=(1−ln x)/x².

Stationary when

1−ln x=0 → ln x=1 → x=e.

For 0<x<e, ln x<1, so f′>0.

For x>e, ln x>1, so f′<0.

Therefore f has a maximum at x=e, with value 1/e.

Solving the derivative equation requires reversing a logarithm, so algebra and calculus remain integrated.

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21 · Verification: let the function family help check the derivative

• Graph direction: e^x and ln x should have positive derivatives on their domains.
• Trig landmarks: sin x has zero gradient at its maxima/minima; cos x has zero gradient where sin x = 0.
• Sign of tan derivative: sec²x should never be negative where tan is defined.
• Chain factor: horizontal compression such as sin(5x) should multiply derivative magnitude by 5 locally.
• Domain: ln(g(x)) derivative cannot be used outside the original log domain.
• Factorisation: e^g(x) never equals zero, so stationary points of products often come from the remaining factors.
• Numerical check: compare a small secant gradient with the derivative value.

Local check for e^x

At x=0, derivative predicts gradient 1. A small secant from 0 to 0.01 has gradient

[e^0.01−1]/0.01 ≈ 1.005.

That is close to 1, supporting the derivative interpretation.

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22 · The advanced differentiation error map

• Degree-mode calculus error: treating standard trig derivative formulas as degree-based rather than radian-based.
• Cosine-sign error: d(cos x)/dx written as +sin x.
• Tangent derivative error: d(tan x)/dx written as sec x instead of sec²x.
• Exponential confusion: d(e^x)/dx written as xe^x−1 by misusing the power rule.
• Logarithm confusion: d(ln x)/dx written as ln′ style or 1/ln x rather than 1/x.
• Chain-factor omission: d[sin(4x)]/dx written as cos(4x) instead of 4cos(4x).
• Nested-order error: e^{sin x} confused with sin(e^x).
• Product-rule omission: derivative of xe^x written only as xe^x.
• Domain loss: derivative of ln(x−3) used for x≤3.
• Stationary exponential error: solving e^x=0 when exponential factors never vanish.
• Trig interval loss: derivative equation solved for only one principal angle.
• Second-derivative sign error: repeated trig derivatives lose minus signs.
• Model unit error: derivative calculated but not interpreted as rate per time unit.

A first weak-link diagnostic

1. Differentiate sin x, cos x, tan x, e^x, ln x from memory.
2. Differentiate sin(3x), e^x², ln(2x+1).
3. Differentiate xe^x and x²ln x.
4. Find one tangent to a trig/exponential/log function.
5. Find and classify a stationary point of xe^−x.
6. Solve a trig stationary-point equation over a stated interval.

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23 · Original guided practice

The questions below are original to this classroom guide. They move from derivative-family recall into mixed-rule calculus and applications.

A · Core derivative families

1. Differentiate 5sin x − 3cos x.
2. Differentiate 4tan x + 2x.
3. Differentiate 7e^x − 4ln x.
4. Find the second derivative of sin x.
5. Find the second derivative of cos x.
6. Find the second derivative of e^x.
7. Find the second derivative of ln x.

B · Chain Rule

8. Differentiate sin(4x−1).
9. Differentiate cos(3x+2).
10. Differentiate tan(5−2x).
11. Differentiate e^6x−3.
12. Differentiate e^x²+1.
13. Differentiate ln(5x+4).
14. Differentiate ln(x²+9).
15. Differentiate sin(x²).
16. Differentiate e^{sin x}.
17. Differentiate sin(e^x).
18. Differentiate ln(cos x) on a domain where cos x>0.

C · Products and quotients

19. Differentiate xsin x.
20. Differentiate x²cos x.
21. Differentiate xe^x.
22. Differentiate x²e^−x.
23. Differentiate xln x.
24. Differentiate x²ln x.
25. Differentiate sin x/x.
26. Differentiate e^x/x.
27. Differentiate ln x/x.
28. Differentiate (1+sin x)/(1+cos x).

D · Tangents and normals

29. Find the tangent to y=e^x at x=0.
30. Find the normal to y=ln x at x=1.
31. Find the tangent to y=sin x at x=π/3.
32. Find the tangent to y=cos x at x=π/2.
33. Find the points on y=e^x where the tangent gradient is 4.
34. Find the x-value where y=ln x has tangent gradient 1/5.
35. Find the points on y=sin x, 0≤x≤2π, where the tangent is horizontal.

E · Stationary points and optimisation

36. Find and classify the stationary point of y=xe^−x.
37. For x≥0, find the maximum value of y=x²e^−x.
38. Find and classify the stationary point of y=ln x−x/2.
39. Find the maximum value of y=ln x/x for x>0.
40. Find and classify the stationary points of y=sin x+cos x for 0≤x≤2π.
41. Find the stationary points of y=2sin x+cos2x for 0≤x≤2π.
42. Find the intervals in 0≤x≤2π where y=sin x is increasing/decreasing.
43. Find the intervals where y=xe^−x is increasing/decreasing.

F · Rates and modelling

44. P(t)=1200e^0.05t. Find P′(t) and express it in terms of P.
45. M(t)=80e^−0.3t. Find M′(t) and interpret the sign.
46. s(t)=4sin(2t). Find velocity and acceleration.
47. A periodic height is h(t)=10+3cos(πt/4). Find dh/dt and the instantaneous vertical rate at t=2.
48. A quantity Q(t)=50e^0.1t. At what time is its instantaneous growth rate 10 units per time?

G · Mixed reasoning

49. Explain why d(sin x)/dx=cos x is tied to radian measure.
50. A student differentiates cos(4x) as −sin(4x). Diagnose the missing factor.
51. A student differentiates e^x² as e^x². Diagnose the missing factor.
52. A student solves e^−x(2−x)=0 and includes a solution from e^−x=0. Explain the error.
53. A student differentiates ln(x²−4) and then uses the derivative at x=0. Explain why this is invalid in the real setting.
54. Explain why a factor e^g(x) can often be ignored when solving a stationary equation after factorisation.
55. Explain why solving cos x−sin x=0 for a stationary-point question still requires interval trigonometry.
56. State the exact G3 derivative-family extension beyond the shared G2 calculus floor.

24 · Worked answers and reasoning checkpoints

1. 5cos x + 3sin x.
2. 4sec²x + 2.
3. 7e^x − 4/x.
4. sin x → cos x → −sin x.
5. cos x → −sin x → −cos x.
6. e^x.
7. −1/x².
8. 4cos(4x−1).
9. −3sin(3x+2).
10. Derivative tan u = sec²u·u′, u′=−2, so −2sec²(5−2x).
11. 6e^6x−3.
12. 2xe^x²+1.
13. 5/(5x+4).
14. 2x/(x²+9).
15. 2xcos(x²).
16. e^{sin x}cos x.
17. e^xcos(e^x).
18. −tan x, on the original domain where cos x>0.
19. sin x+xcos x.
20. 2xcos x−x²sin x.
21. e^x(1+x).
22. xe^−x(2−x).
23. ln x+1.
24. x(2ln x+1).
25. (xcos x−sin x)/x².
26. e^x(x−1)/x².
27. (1−ln x)/x².
28. Let u=1+sin x, v=1+cos x. Derivative = [(1+cos x)cos x+(1+sin x)sin x]/(1+cos x)², which may be simplified further.
29. y=x+1.
30. y=−x+1.
31. y−√3/2=(1/2)(x−π/3).
32. At x=π/2, point (π/2,0), gradient −sin(π/2)=−1. Tangent: y=−(x−π/2).
33. e^x=4 → x=ln4. Point (ln4,4).
34. 1/x=1/5 → x=5.
35. Horizontal tangent when cos x=0: x=π/2,3π/2.
36. y′=e^−x(1−x). Stationary x=1. Sign +→−, so maximum at (1,1/e).
37. y′=xe^−x(2−x). For x≥0, maximum at x=2, value 4/e².
38. y′=1/x−1/2=0 gives x=2. y″=−1/x²<0, so maximum at (2,ln2−1).
39. y′=(1−ln x)/x². Maximum at x=e, value 1/e.
40. Stationary x=π/4,5π/4. Values √2 and −√2. Second derivative gives maximum (π/4,√2) and minimum (5π/4,−√2).
41. y′=2cos x−2sin2x = 2cos x−4sin x cos x = 2cos x(1−2sin x). Stationary where cos x=0 or sin x=1/2. Thus x=π/2,3π/2,π/6,5π/6. Classification can be completed with sign or second derivative.
42. sin x derivative cos x. Increasing where cos x>0: (0,π/2) and (3π/2,2π) within the interval; decreasing where cos x<0: (π/2,3π/2).
43. Derivative sign controlled by 1−x: increasing x<1, decreasing x>1.
44. P′=60e^0.05t=0.05P.
45. M′=−24e^−0.3t=−0.3M. Negative sign means the modelled amount is decreasing.
46. v=8cos2t; a=−16sin2t=−4s.
47. dh/dt=3[−sin(πt/4)](π/4)=−(3π/4)sin(πt/4). At t=2, sin(π/2)=1, so rate −3π/4 height-units per time.
48. Q′=5e^0.1t. Set 10: e^0.1t=2, so t=10ln2.
49. Radians scale angle directly by arc length on the unit circle; that natural scale makes the limiting derivative constant exactly 1 rather than introducing a conversion factor such as π/180.
50. Chain Rule: derivative of 4x is 4. Correct derivative −4sin(4x).
51. Chain Rule: derivative of x² is 2x. Correct derivative 2xe^x².
52. For real x, e^−x>0 and never equals zero. The only stationary solution from the product is x=2.
53. ln(x²−4) requires x²−4>0, so x<−2 or x>2. x=0 is outside the original domain; the derivative is not valid there as the derivative of that real log function.
54. e^g(x) is strictly positive for real g(x), so it contributes no zero and no sign change. Stationary zeros must come from the remaining factor.
55. Trig equations have repeated solutions. The stationary x-values must include every solution of the derivative equation inside the stated interval, not only the principal inverse-trig value.
56. G3 adds derivatives of sin x, cos x, tan x, e^x and ln x to the shared rational-power/Product/Quotient/Chain calculus floor.

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25 · A two-week G3 extension architecture

1. Day 1: five core derivative facts and radian meaning.
2. Day 2: Chain Rule with sin, cos, tan, e^x, ln x.
3. Day 4: Product/Quotient Rule mixed with new functions.
4. Day 6: multi-layer compositions such as e^{sin x} and ln(cos x).
5. Day 8: tangents and normals with exact trig/log/exp values.
6. Day 10: stationary points and increasing/decreasing intervals.
7. Day 12: growth/decay and oscillation rates.
8. Day 14: mixed unlabelled G3 calculus set.

Five-minute retrieval test

• d/dx(sin x)=?
• d/dx(cos x)=?
• d/dx(tan x)=?
• d/dx(e^x)=?
• d/dx(ln x)=?
• Why radians?
• Differentiate sin(3x), e^x², ln(4x+1).

26 · For teachers: the extension should feel like one new row in the function library

Do not restart calculus from zero. Put the new derivative families beside the existing power rule and ask students to run familiar structures through them.

For example, the expression x²(3x+1)⁵ in Chapter 10 had the rule tree Product → Chain. The new expression x²sin(3x) has the same rule tree. Only the derivative of the inner factor has changed.

Likewise, xe^−x is still Product Rule plus Chain Rule. ln(x²+1) is still a nested function. The new chapter should therefore increase transfer rather than create five isolated mini-topics.

Teach radians at the calculus boundary

Students often learn radians earlier as conversion practice and forget why the unit matters. Revisit the meaning now. Calculus gives radians a reason: the standard derivative formulas have their clean form because radian measure is the natural angular scale.

Use graph sense to check symbolic work

If a student writes d(e^x)/dx negative somewhere, the graph contradicts the result. If the derivative of tan x is negative on a regular branch, the graph contradicts the result. If cosine’s derivative lacks a minus sign, inspect the graph near zero. Calculus should not become blind symbol manipulation.

27 · For parents: what real progress sounds like

A student relying on memory may say, “I have five more derivative formulas.” A student gaining structural control begins to say:

• “The outer function is sine, and the inside is 3x, so I need the inner factor 3.”
• “This is Product Rule outside and Chain Rule inside the exponential.”
• “e^−x is always positive, so it cannot create a stationary root.”
• “The log derivative exists only where the original logarithm is defined.”
• “The derivative equation is trigonometric, so I need every solution in the interval.”
• “Radians are the natural angle measure for these derivative formulas.”
• “I can use the same optimisation logic as before; only the function family changed.”

This is the intended progression: the calculus operating system stays stable while the function library grows.

28 · What carries forward

The old textbook’s next major chapter is Integration. SEC 2027 again has a shared G2/G3 floor and a wider G3 function family. Both levels treat integration as reverse differentiation, integrate rational powers and use definite integrals for areas. G3 additionally integrates sin x, cos x, sec²x and e^x, plus corresponding linear-argument forms, and later applies calculus to straight-line kinematics.

The next classroom chapter will therefore begin with the shared integration engine—reverse differentiation, constants of integration, linear-argument power forms, definite integrals and signed area—before the G3-specific integration families are separated cleanly where needed.

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Official syllabus reference

Singapore Examinations and Assessment Board · 2027 SEC G3 Additional Mathematics K341

Curriculum and assessment requirements can change. The official SEAB syllabus remains the controlling source for current subject codes, examinable content and examination structure.

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