Application of Derivatives — chapter notes

If $y = f(x)$, then $\dfrac{dy}{dx}$ (or $f'(x)$) is the rate of change of $y$ with respect to $x$, and $f'(x_0)$ is that rate at $x = x_0$. Examples: velocity $\dfrac{ds}{dt}$; for a circle $\dfrac{dA}{dr} = 2\pi r$; for a sphere $\dfrac{dV}{dr} = 4\pi r^2$.

A positive rate means the quantity is increasing; a negative rate means it is decreasing.

When quantities $x$ and $y$ both depend on time $t$, the chain rule gives $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$.

Method: (1) name the changing quantities; (2) write one relation between them (formula, Pythagoras, similar triangles); (3) differentiate both sides with respect to $t$; (4) substitute the values at the instant asked; (5) answer with units and sign.

Example: a $5$ m ladder, foot moving away at $0.4$ m/s. From $x^2 + y^2 = 25$: $\dfrac{dy}{dt} = -\dfrac{x}{y}\dfrac{dx}{dt} = -\dfrac34(0.4) = -0.3$ m/s when $x = 3$.

For a cone whose shape is fixed, use similar triangles to remove one variable before differentiating.

If $C(x)$ is the total cost of producing $x$ units, the marginal cost is $MC = C'(x)$. If $R(x)$ is the total revenue from selling $x$ units, the marginal revenue is $MR = R'(x)$. Each approximates the cost (or revenue) of producing (or selling) one more unit.

Example: $R(x) = 3x^2 + 10x$ gives $R'(15) = 100$, while $R(16) - R(15) = 103$.

Do not confuse the marginal value $C'(x_0)$ with the total value $C(x_0)$.

$f$ is strictly increasing on $I$ if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ for $x_1, x_2$ in $I$, and strictly decreasing if $x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$.

Test. Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$. If $f'(x) > 0$ on $(a, b)$, $f$ is strictly increasing on $[a, b]$; if $f'(x) < 0$, strictly decreasing; if $f'(x) = 0$, constant.

Method: factorise $f'(x)$, find its zeros, make a sign chart, read off the intervals.

Note: $f(x) = x^3$ is strictly increasing although $f'(0) = 0$: zeros of $f'$ at isolated points are allowed.

$f$ has a local maximum at $c$ if $f(c) \ge f(x)$ for all $x$ in some open interval around $c$ (local minimum: $\le$). Inside an interval, extrema occur only at critical points: where $f'(c) = 0$ or $f'(c)$ does not exist.

First derivative test. As $x$ increases through $c$: $f'$ changes $+$ to $-$: local maximum; $-$ to $+$: local minimum; no change: neither (a point of inflection such as $x^3$ at $0$).

Example: $f(x) = x^3 - 6x^2 + 9x + 1$ has $f'(x) = 3(x - 1)(x - 3)$: local maximum $5$ at $x = 1$, local minimum $1$ at $x = 3$.

Let $f'(c) = 0$ and $f''(c)$ exist.

• $f''(c) < 0$: local maximum at $c$.
• $f''(c) > 0$: local minimum at $c$.
• $f''(c) = 0$: the test fails; use the first derivative test.

Example: $f(x) = x^4 - 8x^2 + 3$: $f''(0) = -16$ (local max $3$), $f''(\pm2) = 32$ (local min $-13$). For $x^4$ at $0$ both $f'$ and $f''$ vanish, yet there is a minimum.

A continuous function on a closed interval $[a, b]$ has an absolute maximum and an absolute minimum. Closed interval method: find the critical points in $(a, b)$, evaluate $f$ there and at $a$ and $b$, and compare.

Example: $x^3 - 27x$ on $[-4, 6]$: values $44, 54, -54, 54$; absolute maximum $54$ (at $-3$ and $6$), absolute minimum $-54$ (at $3$).

Optimisation: draw a figure; write the quantity to optimise; use the condition to reduce it to one variable and state its domain; solve $f'(x) = 0$, rejecting roots outside the domain; justify max or min; answer with units. Example: a $30$ cm square sheet gives the largest open box ($2000$ cm$^3$) with a $5$ cm cut.

Not in the 2026–27 syllabus: tangents and normals, and approximations.