IP AMaths Notes (Upper Sec, Year 3-4): 14) Differentiation Fundamentals

For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.

How this chapter applies

• Eclat core: derivative as tangent gradient and rate of change, standard notation, rational powers, trigonometric, exponential, and logarithmic derivatives form the main route.
• School-sensitive extension: formal limit proofs and deeper first-principles work may vary by school.
• 2027 national comparison: K341 Topic C1 covers derivative meaning, notation, and derivatives of rational powers, sine, cosine, tangent, exponential, and logarithmic functions.
• Check your school: confirm first-principles depth and the required derivative formula set.
• Exam-track route: use the separate O-Level and SEC G3 Additional Mathematics notes for K341 topic ownership.

Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 14) Differentiation Fundamentals cover?
A: First principles, power rule, and basic derivative interpretations for IP AMaths.

Differentiation measures instantaneous rate of change. Know how to move from first principles to the standard rules.

Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.

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The core idea is simple: Differentiation measures instantaneous rate of change.

Use it as a working check: Learn what the derivative means before memorising rules: it gives gradient, tangent slope, and rate of change.

Then go one layer deeper: Example: for y = $x^{2}$, first principles gives gradient 2x; at x = 3, the tangent gradient is 6.

Choosing the differentiation route

Before differentiating, identify what the question is asking for. The same derivative can be used as a formula, a gradient, or a rate of change.

Common trap: substituting the point before differentiating usually gives the value of $y$, not the gradient. Differentiate first when the question asks about slope or rate.

1 Key rules

Limit definition. $f^{\prime}(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$.

Power rule (real $n$, $x>0$). $\dfrac{\mathrm{d}}{\mathrm{d}x}\bigl(x^n\bigr) = nx^{n - 1}$.

For positive integer powers, the rule extends to all real $x$. For negative integer powers, require $x \neq 0$. A constant power $x^0=1$ has derivative $0$.

Constant multiple. $\dfrac{\mathrm{d}}{\mathrm{d}x}\bigl(c \cdot f(x)\bigr) = c \cdot f^{\prime}(x)$.

Sum or difference. $\dfrac{\mathrm{d}}{\mathrm{d}x}\bigl(f(x) \pm g(x)\bigr) = f^{\prime}(x) \pm g^{\prime}(x)$.

Standard functions.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)=\cos x,\quad \frac{\mathrm{d}}{\mathrm{d}x}(\cos x)=-\sin x,\quad \frac{\mathrm{d}}{\mathrm{d}x}(\tan x)=\sec^2x.$$

$$\frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^x)=\mathrm{e}^x,\quad \frac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\frac{1}{x}.$$

The logarithmic rule above requires $x>0$. The tangent rule applies where tangent is defined. Angles in these trigonometric derivative rules are measured in radians. A linear inner function adds its gradient through the chain rule, for example $\dfrac{\mathrm{d}}{\mathrm{d}x}\sin(3x+1)=3\cos(3x+1)$.

Power-rule rewrite checkpoint

Before applying the power rule, rewrite roots and reciprocals as powers of $x$. For the square-root rows below, require $x>0$; for the reciprocal-square row, require $x \neq 0$. This makes the differentiation step mechanical.

Worked check: $y = \dfrac{3}{\sqrt{x}} + \dfrac{2}{x^2}$ becomes $y = 3x^{-\frac{1}{2}} + 2x^{-2}$, so $$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3}{2}x^{-\frac{3}{2}} - 4x^{-3}.$$

Misconception to avoid: the power rule handles fractional and negative powers on their valid domains. Rewriting the expression as a power of $x$ lets you apply the rule directly.

2 Worked example - From first principles

Find $f^{\prime}(x)$ for $f(x) = x^2$ using the limit definition.

1. Compute difference quotient: $\dfrac{(x + h)^2 - x^2}{h} = \dfrac{x^2 + 2xh + h^2 - x^2}{h}$.
2. Simplify: $\dfrac{2xh + h^2}{h} = 2x + h$.
3. Take limit: as $h \to 0$, $f^{\prime}(x) = 2x$.

First-principles cancellation checkpoint

For these polynomial examples, factor and cancel $h$ while $h \neq 0$, then take the limit as $h \to 0$. Do not substitute zero into the original difference quotient.

Worked check: for $f(x) = x^3$,

$$\frac{(x + h)^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2.$$

After the division by $h$, the limit gives $f^{\prime}(x) = 3x^2$.

Misconception check: the step $h \to 0$ is not the first step. It is the final step after expansion, cancellation, and division by $h$.

3 Worked example - Polynomial function

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ for $y = 5x^4 - 3x^2 + 7$.

1. Differentiate term by term: $\dfrac{\mathrm{d}}{\mathrm{d}x}(5x^4) = 20x^3$.
2. $\dfrac{\mathrm{d}}{\mathrm{d}x}(-3x^2) = -6x$; constant term differentiates to $0$.
3. Combine: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 20x^3 - 6x$.

4 Worked example - Equation of a tangent line

Find the equation of the tangent to $y = x^3 - 4x$ at $x = 2$.

1. Differentiate: $\tfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4$.
2. Gradient at $x = 2$ is $3 \times 2^2 - 4 = 8$.
3. Point on the curve: $y = 2^3 - 4 \times 2 = 0$, so the point of tangency is (2, 0).
4. Tangent equation: $y - 0 = 8(x - 2)$, hence $y = 8x - 16$.

Answer: tangent line $y = 8x - 16$.

5 Practice Quiz

Confirm that basic rules, gradients, and tangent setups are second nature before combining them with advanced techniques.

6 Try this

Differentiate $y = \dfrac{3}{\sqrt{x}} + \dfrac{2}{x^2}$ and simplify the result.