Mathematics · 15 / 23
Derivatives: local slope
High school · intermediate · Worked example and 4 exercises with solutions, at your pace.
The principle, simply
A rate of change compares two points. Where it exists, the derivative is the limit as the points approach each other: the tangent slope. The sign of the derivative over an interval tells you how the function varies. The lab’s numerical slope is an approximation, not a proof.
Concept sketch: a tangent’s slope triangle compares vertical and horizontal change. Secant slopes approach the derivative.
- 01Two points
- 02Rate of change
- 03Local limit
- 04Tangent slope
WORKED EXAMPLE
From principle to reasoning
For f(x) = x², f′(x) = 2x. At x = 3 the slope is 6. The tangent is y = f(3) + 6(x − 3) = 6x − 9.
Build the limit for x²
Between x and x+h, nonzero h gives ((x+h)² − x²)/h = (2xh+h²)/h = 2x+h. As h approaches zero, the slope approaches 2x. Never substitute h=0 into the original quotient.
Solve a problem in this subject
- Define the objects
State unknowns, domain and data. Real square roots need nonnegative arguments; denominators must be nonzero.
- Choose a transformation
Name the rule: expansion, equivalent equations, factoring, differentiation. Check its conditions. Squaring an equation requires testing candidates in the original.
- Show the reasoning
Each line should follow from the previous one. Keep exact values where possible and distinguish equality, approximation and implication.
- Test and conclude
Substitute into the original expression. Check graph points and domain boundaries. State a solution set or the required unit.
YOUR TURN
01 · Try before checking
Where does x² decrease and increase?
One hint
Find the sign of 2x.
Reveal and understand the solution
2x is negative for x < 0 and positive for x > 0. The function decreases to zero, then increases, with a minimum at zero.
Compare method, conditions and answer; identify the exact part to revisit.
IMPROVE THROUGH PRACTICE
Three additional challenges with solutions
Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.
Differentiate f(x) = 3x² − 4x + 7.
A hint to get started
Differentiate each term separately.
Step-by-step solution
- The derivative of 3x² is 6x.
- The derivative of −4x is −4, and of 7 is zero.
- f′(x) = 6x − 4.
Which step did you miss? Retry without help, then explain why the method works.
Find the tangent to f(x) = x² at x = 2.
A hint to get started
You need both a point and a slope.
Step-by-step solution
- f(2) = 4 and f′(2) = 4.
- y = 4(x − 2) + 4.
- The tangent is y = 4x − 4, not y = 4x.
Which step did you miss? Retry without help, then explain why the method works.
Study the variation of f(x) = x³ − 3x.
A hint to get started
Factor the derivative to find its sign.
Step-by-step solution
- f′(x) = 3(x − 1)(x + 1).
- It is positive outside [−1,1] and negative between the endpoints.
- f increases, decreases, then increases; f(−1) = 2 and f(1) = −2.
Which step did you miss? Retry without help, then explain why the method works.
The trap to avoid
A zero derivative at one point does not alone prove a maximum; inspect neighbouring signs.
Without notes, can you…
- Explain the principle in your own words?
- Rebuild the diagram and its connections?
- Solve the exercise and justify the method?
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