The Derivative as an Instantaneous Rate of Change
The derivative function, denoted $f'(x)$ or $\frac{dy}{dx}$, gives the exact slope (gradient) of the tangent line to the curve $y = f(x)$ at any point $(x, y)$:
In physical models, if $s(t)$ represents displacement, $v(t) = s'(t)$ represents instantaneous velocity; if $C(x)$ is cost, $C'(x)$ is marginal cost.
The Power Rule for Differentiation
For any term $ax^n$ where $n \in \mathbb{Q}$:
Always convert fractional and negative exponents into power form before differentiating:
- $\frac{1}{x} = x^{-1} \implies \frac{d}{dx}(x^{-1}) = -1x^{-2} = -\frac{1}{x^2}$
- $\frac{4}{x^3} = 4x^{-3} \implies \frac{d}{dx}(4x^{-3}) = -12x^{-4} = -\frac{12}{x^4}$
- $\sqrt{x} = x^{0.5} \implies \frac{d}{dx}(x^{0.5}) = 0.5x^{-0.5} = \frac{1}{2\sqrt{x}}$
- The derivative of a constant term is zero: $\frac{d}{dx}(c) = 0$.
Equations of Tangents and Normals
At a specific point $(x_1, y_1)$ on the curve:
- Tangent Line: Shares the exact gradient of the curve: $m_t = f'(x_1)$.
$$y - y_1 = m_t(x - x_1)$$
- Normal Line: Perpendicular to the tangent line at $(x_1, y_1)$: $m_n = -\frac{1}{m_t}$.
$$y - y_1 = -\frac{1}{m_t}(x - x_1)$$
Reading and estimating rates
- The derivative has units of output per input. Positive derivative means increasing value, not necessarily positive value. A zero derivative need not be a turning point.
- Estimate a rate at t using the symmetric secant $[f(t+h)-f(t-h)]/(2h)$. It is an estimate from finite data, not an exact instantaneous rate.
- If f′(a)=0, the tangent is horizontal y=f(a) and the normal is vertical x=a. A cusp or discontinuity can make the derivative undefined.
Given $f(x) = 3x^3 - 5x^2 + 7x - 12$, find $f'(x)$.
Solution:
Differentiate term by term using the power rule: [M1]
$\frac{d}{dx}(3x^3) = 9x^2$, $\frac{d}{dx}(-5x^2) = -10x$, $\frac{d}{dx}(7x) = 7$, $\frac{d}{dx}(-12) = 0$. [A1]
$f'(x) = 9x^2 - 10x + 7$. [A1]
Find the gradient of the curve $y = 2x^2 - 8x + 3$ at the point where $x = 3$.
Solution:
$\frac{dy}{dx} = 4x - 8$. [A1]
Substitute $x = 3$: $\frac{dy}{dx}\Big|_{x=3} = 4(3) - 8 = 12 - 8$ [M1]
Gradient $m = 4$. [A1]
Find the equation of the tangent line to the curve $y = x^2 - 4x + 1$ at the point $P(3, -2)$. Give your answer in the form $y = mx + c$.
Solution:
$\frac{dy}{dx} = 2x - 4$. At $x = 3$: $m_t = 2(3) - 4 = 2$. [M1][A1]
Tangent line through $(3, -2)$ with slope $2$: $y - (-2) = 2(x - 3)$ [M1]
$y + 2 = 2x - 6 \implies y = 2x - 8$. [A1]
Find the equation of the normal line to the curve $y = x^2 - 4x + 1$ at the point $P(3, -2)$.
Solution:
Tangent gradient at $x = 3$ is $m_t = 2$.
Normal gradient: $m_n = -\frac{1}{m_t} = -\frac{1}{2} = -0.5$. [M1][A1]
Equation of normal: $y - (-2) = -0.5(x - 3)$ [M1]
$y + 2 = -0.5x + 1.5 \implies y = -0.5x - 0.5$ (or $x + 2y + 1 = 0$). [A1]
Differentiate $g(x) = \frac{4}{x^2} + 5\sqrt{x}$ with respect to $x$.
Solution:
Rewrite with index notation: $g(x) = 4x^{-2} + 5x^{0.5}$. [M1]
$g'(x) = 4(-2)x^{-3} + 5(0.5)x^{-0.5}$ [M1]
$g'(x) = -8x^{-3} + 2.5x^{-0.5} = -\frac{8}{x^3} + \frac{2.5}{\sqrt{x}}$. [A1]
Find the coordinates of the point on the curve $y = 2x^2 - 10x + 7$ where the tangent is horizontal.
Solution:
A horizontal tangent has gradient $m = 0 \implies \frac{dy}{dx} = 0$. [R1]
$\frac{dy}{dx} = 4x - 10 = 0 \implies 4x = 10 \implies x = 2.5$. [M1][A1]
Find $y$: $y = 2(2.5)^2 - 10(2.5) + 7 = 2(6.25) - 25 + 7 = 12.5 - 25 + 7 = -5.5$. [M1]
Coordinates: $(2.5, -5.5)$. [A1]
A particle's displacement is $s(t) = t^3 - 6t^2 + 9t + 4$ meters. Find its velocity when $t = 4\text{ seconds}$.
Solution:
Velocity $v(t) = s'(t) = 3t^2 - 12t + 9$. [M1][A1]
When $t = 4$: $v(4) = 3(4^2) - 12(4) + 9 = 3(16) - 48 + 9 = 48 - 48 + 9 = 9\text{ m/s}$. [A1]
Show that the curve $y = x^3 + 3x$ has no points where the gradient is negative, and deduce whether the function is strictly increasing.
Solution:
$\frac{dy}{dx} = 3x^2 + 3$. [M1][A1]
Since $x^2 \ge 0$ for all real $x$, $3x^2 \ge 0$, which means $3x^2 + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$. [R1]
Since $\frac{dy}{dx} > 0$ everywhere, the function is strictly increasing. [A1]
Find the points on the curve $y = \frac{1}{3}x^3 - 4x + 5$ where the tangent is parallel to the line $y = 5x - 7$.
Solution:
Parallel tangents must have the same slope: $m = 5$. [R1]
$\frac{dy}{dx} = x^2 - 4$. Set $x^2 - 4 = 5 \implies x^2 = 9 \implies x = \pm 3$. [M1][A1]
When $x = 3$: $y = \frac{1}{3}(27) - 4(3) + 5 = 9 - 12 + 5 = 2 \implies (3, 2)$.
When $x = -3$: $y = \frac{1}{3}(-27) - 4(-3) + 5 = -9 + 12 + 5 = 8 \implies (-3, 8)$. [A1]
Explain the geometric relationship between the tangent line and the normal line at any point on a smooth curve.
Solution:
The tangent line touches the curve locally and shares the curve's instantaneous direction. [A1]
The normal line passes through the point of contact perpendicular ($90^\circ$) to the tangent line. [A1]
Their gradients satisfy $m_{\text{tangent}} \times m_{\text{normal}} = -1$ (provided neither line is horizontal or vertical). [R1]
Tank volumes at minutes4,5,6are920,900,876litres. Estimate the rate at5minutes and compare the one-sided rates.
Central estimate(876−920)/2=−22L/min. One-sided estimates−20and−24L/minsuggest drainage is accelerating. Measurement spacing limits precision.
For f(x)=x³ find the tangent and normal at0. Is this a local extremum?
f′0=0andf0=0: tangent y0,normalx0. f′=3x²ispositive on both sides, so0is a stationary inflection, not an extremum.
Core Formulas (Topic 5.1 - 5.3)
| Power Rule | $$\frac{d}{dx}(a x^n) = a n x^{n-1}$$ |
| Tangent Line Equation | $$y - y_1 = m_t(x - x_1) \quad \text{where } m_t = f'(x_1)$$ |
| Normal Line Equation | $$y - y_1 = -\frac{1}{m_t}(x - x_1)$$ |
| Horizontal Tangent Condition | $$f'(x) = 0$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Forgetting that $\frac{d}{dx}(\text{constant}) = 0$: Writing $\frac{d}{dx}(5) = 5$ is a common careless slip.
- Trap 2: Differentiating denominators directly: In $\frac{1}{x^2}$, you must write $x^{-2}$ and get $-2x^{-3} = -\frac{2}{x^3}$. Do not write $\frac{1}{2x}$!
- Trap 3: Tangent vs Normal Slope confusion: Tangent gradient is $m$, normal gradient is $-\frac{1}{m}$. Always double-check which line the question asked for.
- Trap 4: Forgetting the $y$-coordinate: If given only the $x$-value, substitute $x$ into the original function $f(x)$ to get $y_1$ (not into $f'(x)$).
- Trap 5: Using degrees mode on GDC for rates of change: While AI SL differentiation covers polynomials, ensure angles are always in radians when calculus meets trigonometry.
5-Question Concept Check
Score: - / 51. What is the derivative of $f(x) = 5x^4 - 6x + 9$?
2. What is the gradient of the tangent to $y = x^3 - 2x$ at $x = 2$?
3. If the tangent gradient to a curve is $4$, what is the gradient of the normal line at the same point?
4. What is $\frac{d}{dx}\left(\frac{1}{x}\right)$?
5. At what value of $x$ does the curve $y = x^2 - 6x + 8$ have a horizontal tangent?