Higher: use derivatives for gradients and rates
Evaluate and solve derivative conditions, return to original coordinates, and filter rate answers through the stated model domain.
Exam scope and assessment notes
Higher only: Foundation has no section3.4 Calculus requirement. Higher also includes all shared Foundation content. These six original lessons teach variable/local versus interval rates, termwise integer-power differentiation, gradients/rates, stationary coordinates and graph-shape classification, one-dimensional kinematics and other simple practical models. H:3.4C requires both the gradient/rate and stationary-point lessons; H:3.4E requires both the motion and other-practical lessons with worked examples and all Guided, Independent and Transfer stages. The six-lesson split is StudyForA’s teaching design, not a syllabus-prescribed count. Small negative-integer powers with x≠0 conservatively retain integer-power breadth; fractional/trigonometric differentiation, integration and product/quotient/chain rules are not inferred requirements. H:3.4D classifies maxima and minima by general original-graph shape only; a second derivative is used for acceleration in motion, not as a substitute classification test. A derivative zero supplies an input, not the original stationary ordinate or an automatic turning point. Keep every original domain, physical restriction, admissible integer count and endpoint. Continuous relaxation supports differentiation for discrete counts; it does not make noninteger production feasible. Accepted boundary maxima need not be stationary. The separate H3.3D drawn-tangent construction is not replaced or newly credited by calculus. Fixed interactive examples support exact algebraic self-checks and approximate graph exploration; no general expression/root solver or written-work marker is supplied. Written work is not automatically marked. Attempt acknowledgements and revealed references do not award marks, prove a construction or establish mastery. All five Calculus source rows have structural teaching and staged-practice links; exhaustive subskill approval, specialist approval and cumulative formal assessment remain unestablished. The wider Maths A course and other Pearson qualifications remain incomplete. Publisher modular-map tags and uncorrected proof navigation are separate evidence: Book2 Graphs9 map pages386–415 differs from proof start385 and its later calculus subheadings. Full final-book and Teaching Hub body/pagination alignment remains unverified; chapter links confer no full-book or modular teaching credit.
Before you begin
- Differentiate sums of integer powers with exact coefficients.
- Solve simple quadratic equations and retain signed roots.
- Read original and rate graphs, with physical units and domains.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation finds inputs where a curve f has gradient 6?
What you will learn
- Find a curve gradient by evaluating its derivative at a specified allowed input.
- Solve a prescribed-gradient equation and recover every corresponding original point.
- Interpret practical derivative units and reject candidate inputs outside the model domain.
f(x)=x³ − 4x + 2. Original domain: x in all real inputs. Original values use output units; first rates use output units per input unit.
The panels share the same numbered horizontal scale and have independent vertical scales. Original height, derivative value and derivative zero have different meanings. No input, derivative solution or reference point is preselected.
No input is preselected. Choose a point on your own drawing or a rate input you want to check.
Blue: original; purple: first derivative. Vertical scales differ. Filled boundary markers are included; open markers show excluded finite limits or holes. Traces split at original exclusions and are clipped to each panel; offscreen points are omitted rather than moved to a border. A wider view does not change a domain.
A derivative value is a gradient, not a height
For y=f(x), first differentiate to obtain f′(x). At input c, the original point is (c,f(c)), while its local gradient is f′(c). On a separate derivative graph the point (c,f′(c)) records the rate. Keep those two vertical meanings distinct when marking the graphs.
A positive derivative value means the original curve rises locally, a negative value means it falls, and zero means a horizontal local direction at a differentiable point. If useful, combine the gradient m and original point (c,f(c)) in the known line equation y−f(c)=m(x−c). This is an algebraic check; it does not replace drawing and measuring a tangent in the separate graphical lesson.
Solve the rate condition and keep every candidate
To find points with gradient k, solve f′(x)=k rather than f(x)=k. Factor or rearrange using known algebra, retaining both signed solutions of a square equation when both are permitted. Then substitute every accepted input into the original function to obtain complete coordinate pairs.
A derivative equation can have several solutions, one solution or no permitted solutions. Do not cancel a factor that could be zero or discard a negative input simply because it is negative. Check the original domain first: a negative algebraic input can be valid for a polynomial curve but invalid as a time in a stated practical interval.
Rates use the model’s units and domain
If V is volume in litres and t is time in hours, dV/dt has units litres per hour. A selected rate is not itself the volume, and the rate graph’s height must not be copied as the original volume coordinate. Draw the original model and its rate graph with their own vertical labels and the same selected input marks.
Keep the time interval used to define the model. A derivative polynomial can be evaluated algebraically outside it, but those values do not become valid model times. At an endpoint, interpret any reported rate from the permitted side of the model, consistent with the polynomial extension. A cropped view changes what is seen, not which mathematical or physical inputs are allowed.
Sources for this lesson
Pause and explain
Solving a derivative condition gives times −2 h and 4 h, but the model is defined only for 0≤t≤3 h. Which time is an admissible answer?
Worked example
For f(x)=x³−4x+2, find the point and gradient at x=2. Mark the original point and the corresponding point on a separately labelled derivative graph, showing how the gradient relates to the original curve. Optionally use the known point-and-gradient line method to write a checking tangent equation. The domain is all real x.
Show the worked solution
- Differentiate term by term: f′(x)=3x²−4. Evaluate the rate at the chosen input: f′(2)=3·4−4=8, so the original curve has positive local gradient 8 there.
- Return to the original function for its height: f(2)=8−8+2=2. The original point is (2,2), while the derivative graph records (2,8). Mark these separately and draw the rising local direction on the original cubic.
- As an optional algebraic check, the line through (2,2) with gradient 8 satisfies y−2=8(x−2), giving y=8x−14. This exact derivative-based calculation is separate from a freehand tangent estimate. Both original and derivative polynomial domains are all real x.
Answer f′(x)=3x²−4; at x=2 the original point is (2,2) and the gradient is 8. Derivative point: (2,8). Optional checking tangent: y=8x−14.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For f(x)=2x³−3x²+1, find the original point and gradient at x=−1. Show the point on the original curve, the rate on a separately labelled derivative graph and the local tangent direction. Write a checking line through the original point using its gradient. State the domains.
Give me a hint
Compute f(−1) and f′(−1) separately, then use the original height in the point-and-gradient line equation.
Compare my reasoning
- Differentiate: f′(x)=6x²−6x. At x=−1 its value is 6+6=12, so the original curve rises locally with gradient 12.
- The original height is f(−1)=2(−1)³−3(−1)²+1=−2−3+1=−4. Mark (−1,−4) on the original curve and (−1,12) on the derivative graph.
- The checking line satisfies y+4=12(x+1), hence y=12x+8. Both domains are all real x. The negative input is valid; the negative original height is not a negative gradient.
Original point (−1,−4), gradient 12; f′(x)=6x²−6x. Checking tangent y=12x+8. Both domains: all real x.
Look for these in your work
- Differentiate correctly and evaluate the signed input.
- Keep original height and derivative height separate.
- Use the original point with its rate to check the local tangent line.
Find every point on f(x)=x³−3x+4 whose gradient is 9. Show all accepted points on the original cubic and their corresponding rate marks on a separate derivative graph. Retain signed solutions and return to the original function for every ordinate; optional checking tangent equations may be added. The domain is all real x.
Give me a hint
Set the derivative equal to the requested gradient. A square equation can give two signed inputs; substitute both into the original function after checking the domain.
Compare my reasoning
- Differentiate to get f′(x)=3x²−3. Solve 3x²−3=9, giving x²=4 and x=−2 or x=2. Both are in the all-real domain.
- The original heights are f(−2)=−8+6+4=2 and f(2)=8−6+4=6. Thus the required original points are (−2,2) and (2,6), while the derivative points are (−2,9) and (2,9).
- Draw the original and derivative graphs with these corresponding inputs labelled. Both original local directions have gradient 9 but start at different heights. Optional checking lines are y−2=9(x+2), or y=9x+20, and y−6=9(x−2), or y=9x−12.
Required points: (−2,2) and (2,6). Derivative f′=3x²−3 equals 9 at both. Optional checking tangents: y=9x+20 and y=9x−12.
Look for these in your work
- Solve the derivative condition and retain both signed inputs.
- Recover and correctly pair both original ordinates.
- Show original and derivative meanings separately and check the same prescribed gradient at both points.
An invented tank-volume model is V(t)=t³−3t²+12t+10 litres for 0≤t≤4 hours. Find when its filling rate is 21 L/h, reject any candidate outside the model and give the volume then. Show the accepted input on the original and rate graphs. Also identify the minimum rate over the model interval and use it to check whether this model keeps filling.
Give me a hint
Differentiate volume with respect to hours, solve the rate equation, and apply the time interval before calculating a physical answer. Complete the rate square or inspect its upward graph for its minimum.
Compare my reasoning
- The filling-rate function is V′(t)=3t²−6t+12 L/h. Setting it equal to 21 gives 3t²−6t−9=0, or (t−3)(t+1)=0, so candidate times are t=3 and t=−1.
- Only t=3 h belongs to 0≤t≤4. The original model then gives V(3)=27−27+36+10=46 L. Mark (3,46) on the volume graph and (3,21) on the rate graph; the rejected t=−1 does not become a physical time just because its algebraic rate also equals 21.
- The rate can be written V′(t)=3(t−1)²+9, an upward parabola with minimum 9 L/h at t=1 h, inside the interval. The rate is therefore positive throughout the model, consistent with continuing to fill. Keep litres, hours and litres per hour on the correct axes and do not extrapolate beyond the stated domain.
Accepted time 3 h, volume 46 L; t=−1 h is rejected. Rate V′=3t²−6t+12 L/h has minimum 9 L/h at 1 h and remains positive on 0≤t≤4.
Look for these in your work
- Solve the rate condition and reject the out-of-domain time.
- Return to the original volume and state its units.
- Separate the original/rate graph labels and check the minimum rate within the model interval.
Common mistakes
- Solving f(x)=k when k is a requested gradient rather than an original height.
- Returning a rate instead of the original ordinate, or losing a second signed solution.
- Accepting an out-of-domain model time or confusing a derivative-based tangent check with the separate graphical construction.
What can you explain now?
At input x=−2, a curve has f(−2)=7 and f′(−2)=−5. What are its original point and local gradient?
Compare with the explanation
Original point (−2,7); local gradient −5.
The original function supplies the height 7, while the derivative supplies the signed rate −5. The corresponding derivative-graph point would be (−2,−5).
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Make it stick
Describe the sequence differentiate, solve or evaluate, check the domain, return to the original height. Explain why an original point and a derivative point at the same input have different meanings.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.