Applications of differentiation

First derivative test near a stationary input a

Here $a^-$ and $a^+$ mean inputs just to the left and right, not powers. All source cases and tangent directions are retained:

Case 1

• f′ left: positive.
• f′ at a: zero.
• f′ right: negative.
• Tangent directions: rising → horizontal → falling.
• Source classification: Local maximum.

Case 2

• f′ left: negative.
• f′ at a: zero.
• f′ right: positive.
• Tangent directions: falling → horizontal → rising.
• Source classification: Local minimum.

Case 3, increasing

• f′ left: positive.
• f′ at a: zero.
• f′ right: positive.
• Tangent directions: rising → horizontal → rising.
• Source classification: Stationary inflection in the source’s sketch.

Case 3, decreasing

• f′ left: negative.
• f′ at a: zero.
• f′ right: negative.
• Tangent directions: falling → horizontal → falling.
• Source classification: Stationary inflection in the source’s sketch.

For the last two cases, confirm the concavity change as explained above. The mathematical reconstructions below show the source’s local configurations, including both stationary-inflection orientations.

Second derivative test at a stationary input

Assume the usual differentiability conditions near a and $f^{\prime}(a)=0$. The complete source test has three outcomes:

• Negative f″ at a: Local maximum.

• Positive f″ at a: Local minimum.

• Zero f″ at a: Inconclusive; use another test such as the first derivative test.

A zero second derivative is not proof of an inflection. The source recommends the first derivative test if the second derivative is tedious. Example 1 retains both analytical tests and the separate numerical calculator workflow.

Test a stationary point when the second derivative is zero

For $f(x)=x^3$ and $g(x)=x^4$, both first and second derivatives vanish at zero. Classify each stationary point using behaviour on both sides. Explain why a zero second derivative cannot supply the classification on its own.

After your attempt: compare the reasoning

For $f$, $f^{\prime}(x)=3x^2$ is positive on both sides of zero, so there is no local extremum. Also $f^{\prime\prime}(x)=6x$ changes sign, giving a stationary inflection. For $g$, $g^{\prime}(x)=4x^3$ changes from negative to positive, giving a minimum; $g^{\prime\prime}(x)=12x^2$ has no concavity sign change. The second derivative test is inconclusive when its value is zero.

Concavity describes how the slope changes

Read the two derivatives separately: $f^{\prime}$ describes whether the function rises or falls; $f^{\prime\prime}$ describes whether its slope increases or decreases. A curve can fall while its slope increases.

The source defines the second derivative as $d(dy/dx)/dx=d^2y/dx^2$: the first derivative is the rate of change of y with respect to x; the second is the rate of change of that first derivative with respect to x.

If $f^{\prime\prime}(x)>0$ throughout an interval I, then the slope $f^{\prime}$ is increasing and f is concave upwards there. The page 35 illustrations show tangents below the curve, both for a decreasing curve on (a,b) and an increasing curve on (c,d).

If $f^{\prime\prime}(x)<0$ throughout I, the slope decreases and f is concave downwards: the tangents lie above the source’s increasing curve on (a,b) and decreasing curve on (c,d).

These are interval statements under the usual differentiability assumptions; they are not decided from a single tangent or from whether the function is above the x-axis.

Compare the four cases in the complete page 36 graphical summary:

Source case 1

• f′: positive.
• Function behaviour: Strictly increasing.
• f″: negative.
• Concavity: Downwards.

Source case 2

• f′: positive.
• Function behaviour: Strictly increasing.
• f″: positive.
• Concavity: Upwards.

Source case 3

• f′: negative.
• Function behaviour: Strictly decreasing.
• f″: negative.
• Concavity: Downwards.

Source case 4

• f′: negative.
• Function behaviour: Strictly decreasing.
• f″: positive.
• Concavity: Upwards.

The source asks whether strict increase implies $dy/dx>0$. The Teacher supplies No, with counterexample $y=x^3$: it is strictly increasing but has derivative zero at 0. This reinforces the earlier sufficient-test qualification, not a converse theorem. Example 4 preserves the source’s shuffled interval choices and Teacher matching arrows, while distinguishing rounded graph labels from exact derivative boundaries.

Inflection means a change of concavity

The source compares $y=x^3$ with $y=x^3+x$. Both change from concave down for $x<0$ to concave up for $x>0$ at $(0,0)$, and both have second derivative zero there.

For $x^3$, the first derivative is zero, so the inflection is stationary. For $x^3+x$, the first derivative is 1, so it is non-stationary; the source explicitly labels non-stationary inflection as outside its H2 syllabus scope. An inflection is not a local maximum or minimum in these configurations.

The source says an inflection has $f^{\prime\prime}=0$, but the converse fails. Its counterexample is $y=x^4$: $f^{\prime\prime}(0)=0$ while the curve stays concave upwards on both sides. The existence condition matters: if the second derivative exists at an inflection in this setting, it is zero.

Without that condition, $f(x)=x|x|$ changes concavity at 0 but has no second derivative there. This last example is an independently authored qualification, not a change to the retained source statement.

Sketch the function and derivative together

Page 38 asks for $Y_1=0.1X(X-6)(X+6)$ and its derivative using a G.C. The source callout shows $Y_2$ as the numerical derivative of $Y_1$ with respect to X, evaluated at the current X: use MATH → 8 for the derivative template and the illustrated ALPHA → TRACE → ENTER route to insert $Y_1$.

The printed graphs have original turning labels approximately $(-3.46,8.31)$ and $(3.46,-8.31)$, and derivative x-intercepts approximately $-3.46,3.46$. The Teacher fills the answer: turning points of f correspond to x-intercepts of f′.

Here the analytical derivative is $0.3x^2-3.6$, with exact zeros $\pm2\sqrt3$; a numerical derivative display is an approximation.

Both source QR codes point to the function-and-derivative exploration. This is a separately decoded source pointer; its external content has not been reviewed.

The source’s four-step derivative-sketch procedure

1. Inspect asymptotes. The printed table maps vertical $x=a$ to vertical $x=a$, horizontal $y=a$ to horizontal $y=0$, and oblique $y=ax+b$ to horizontal $y=a$ for the derivative graph.
2. Inspect stationary points. A maximum at (a,b) gives a derivative zero (a,0) crossed positive-to-negative; a minimum gives negative-to-positive. A decreasing stationary inflection gives a derivative local maximum touching(a,0) from below; an increasing stationary inflection gives a derivative local minimum touching(a,0) from above.
3. Inspect slope signs. Negative original gradient puts the derivative below the x-axis; positive original gradient puts it above.
4. Combine steps 1–3 into the sketch, preserving the original domain and every separate branch.

The source asymptote table is a sketching guide for the regular curves being considered, not a theorem for every differentiable function. Check the actual slope behaviour before transferring asymptotes.

For instance, on $x>0$, $f(x)=\sin(x^2)/x$ approaches horizontal asymptote 0, but $f^{\prime}(x)=2\cos(x^2)-\sin(x^2)/x^2$ does not approach 0. Adding $ax+b$ gives the same warning for an oblique asymptote; vertical asymptotes also require inspecting one-sided slopes. These qualifications are independently authored. The paired local sketches below preserve all four stationary-point correspondences.

Explain what the derivative graph cannot locate

Compare $f(x)=x^3-3x$ with $g(x)=x^3-3x+5$. Predict how their function graphs differ and how their derivative graphs compare. Which stationary-point information can the derivative graph determine, and what extra information is needed for the heights?

After your attempt: compare the reasoning

Both derivatives are $3x^2-3$. Both functions increase for $x<-1$ and $x>1$, and decrease for $-1<x<1$. Each has a local maximum at input $-1$ and a local minimum at input $1$.

For $f$, the corresponding heights are 2 and $-2$; for $g$, they are 7 and 3.

A vertical translation leaves slopes unchanged, so a derivative graph alone does not fix these heights. A value of the function at one input would fix the translation in this comparison.

Tangents and normals through a specified point

For $P(x_1,y_1)$ on $y=f(x)$, the tangent has the curve’s local direction at P; the normal passes through P perpendicular to that tangent. The source calls the tangent a line touching the curve. This does not mean that a tangent cannot cross the curve.

Finite tangent slopes

First check that the specified point lies on the curve. When $m=f^{\prime}(x_1)$ is finite, use the point and tangent slope:

$y-y_1=m(x-x_1).$

If $m\ne0$, the normal has slope $-1/m$:

$y-y_1=-\frac{x-x_1}{m}.$

The source remark gives equal slopes for parallel finite-slope lines, and negative reciprocal slopes for perpendicular lines.

Always substitute the specified point after finding the slope.

Horizontal and vertical tangents

The reciprocal formula requires a nonzero finite tangent slope. A horizontal tangent ($m=0$) has vertical normal $x=x_1$.

A vertical tangent, when it exists, has horizontal normal $y=y_1$; it is not represented by a finite derivative value. These special-case qualifications are independently authored.

Connect the method to the examples

Example 8 preserves all six printed calculator steps across pages 43–44, with an independent exact derivative check.

Self-Practice 2 adds analytical working to the source’s final-answer-only exercise.

Example 9 preserves the complete implicit differentiation and negative-reciprocal normal calculation.

Example 7 retains the preceding derivative sketch and its lecture attribution; it repeats Review 4, whose printed 3 marks do not appear in the lecture.

Connect rates using the chain rule

If $y=f(x)$ and both x and y vary with time t, $dx/dt$ is the rate of change of x with respect to time, and $dy/dt$ is the rate of change of y with respect to time. The source connects them by the chain rule:

$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}.$

Differentiate before substituting the instant

Use this order:

1. Name the changing variables and their units.
2. Write an equation connecting them and eliminate extra variables if needed.
3. Differentiate with respect to time.
4. Substitute the values at the requested instant, then state the rate with its sign and units.

A given value such as $r=5$ is an instant’s radius, not a statement that r stays constant.

Keep units on the final rate.

Example 12 preserves both source methods: the Teacher’s completed chain-rule method and the implicit time-differentiation method already printed in both copies.

Choose lengths from the geometry

For triangle areas in Examples 10–11, use intercepts to find the base and height, rather than substituting a line’s slope directly into an area formula. Example 10’s base is the distance between two x-intercepts; Example 11’s base and height are the positive axis intercepts. Exact surds and radians are retained.

Connect the volume rate to the changing depth

A fixed conical tank has height 12 cm and top radius 6 cm. Water enters at $2\ \mathrm{cm}^3/\mathrm{s}$. Find the rate of increase of water depth when it is 4 cm. Predict how this rate changes at depth 8 cm, with the same inflow, and justify the comparison.

After your attempt: compare the reasoning

Similar triangles give $r=h/2$ for the water surface, so

$V=\frac13\pi r^2h=\frac{\pi h^3}{12}.$

Differentiate the relation before substituting the depth:

$\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.$

At $h=4$, $dh/dt=1/(2\pi)\ \mathrm{cm}/\mathrm{s}$. At $h=8$, it is $1/(8\pi)\ \mathrm{cm}/\mathrm{s}$, one quarter as large. The surface area through which the level rises is proportional to $h^2$; the volume rate stays fixed while the depth rate changes.

Formulate an optimisation problem before differentiating

The source’s complete five-step procedure is:

1. Identify variables; if useful, draw a diagram with all given information labelled.
2. Form equations relating those variables.
3. Use all constraints to express the quantity to optimise in a single variable.
4. Differentiate to find stationary values.
5. Test their nature with the first or second derivative test, then answer the original question.

The source describes this as both formulating and solving the mathematical model. Its cone example recommends a sketch to visualise the changing dimensions; its capped-container “Thinking Process” works backwards from the goal, then presents the volume relation, reduced area, stationary calculation and minimum test in a logical order. These are part of the worked steps, rather than a separate teacher-guide section.

Check the feasible domain and boundaries

Independently, restrict the variable to its physically feasible domain and check boundary behaviour before claiming an overall optimum.

A stationary point alone does not establish a global maximum or minimum.

A positive second derivative is sufficient at a stationary point but is not necessary for every minimum; the source also permits a first derivative test.

Count exposed surfaces carefully: a hemisphere-capped cylinder has one flat base, while the source’s ordinary closed can has two ends.

Connect the model to the examples

Example 13 distinguishes the negative height derivative from the positive rate of dropping. Examples 14–16 retain all printed optimisation working, source planning prompts, angle restrictions, sign-table directions and source diagrams. The can calculation also answers the opening question: for the closed-can model, optimal height equals diameter.

Use the mastery checklist to review each skill after lecture and again after tutorial. The practice map below links the printed question references to their worked solutions.

Compare a stationary value with the feasible boundaries

A rectangular region has area $A=x(18-x)$, with a design restriction $2\le x\le6$ metres. Find the greatest feasible area. Explain why solving $A^{\prime}=0$ is not enough.

After your attempt: compare the reasoning

The stationary input is 9, outside the feasible interval. On $2\le x\le6$, $A^{\prime}=18-2x>0$, so the area increases throughout. The largest feasible area is $A(6)=72\ \mathrm{m}^2$ at the boundary. In this model a global optimum can occur at an allowed endpoint even though the derivative there is not zero.