Rates, inputs and tangents
Use the derivative for slope and the original function for height; preserve every domain restriction
\[y-f(a)=f\prime(a)(x-a)\]
- A rate at an input Differentiate the rule, then substitute the input into the derivative. Use the original rule for a point height \(\displaystyle m=f\prime(a),\quad y_0=f(a)\)
- An input for a rate Set the derivative equal to the required rate, solve exactly, then filter against the original domain \(\displaystyle f\prime(x)=m\)
- Stationary inputs Set the derivative equal to zero. Repeated roots give one distinct input. Zero gradient alone does not tell you whether the curve turns
- Tangent through the curve point Use the derivative slope together with the original function value. Rearrange only after forming the point-slope equation \(\displaystyle y-f(a)=f\prime(a)(x-a)\)
- Average and instantaneous rates An average uses two endpoints; an instantaneous rate uses the derivative at one input. Their units agree but their values need not \(\displaystyle \text{average}=\frac{f(b)-f(a)}{b-a},\quad\text{instantaneous at }a=f\prime(a)\)
Worked example
Find the tangent to the curve at the stated input
\[f(x)=x^3-x+2,\quad a=-1\]
- \(\displaystyle f\prime(x)=3x^2-1,\quad f\prime(-1)=2\)Differentiate and evaluate to find the tangent slope
- \(\displaystyle f(-1)=(-1)^3-(-1)+2=2\)Use the original polynomial for the curve point
- \(\displaystyle y-2=2(x+1)\)The tangent passes through (−1, 2) with slope 2.
- \(\displaystyle y=2x+4\)The tangent intercept is 4. Substituting x = −1 gives y = 2 as required
Common mistakes
Solving the wrong equation
✕\(\displaystyle f(x)=m\)✓\(\displaystyle f\prime(x)=m\)
A specified gradient is a condition on the derivativeUsing the derivative as point height
✕The tangent passes through (a, f′(a))✓The tangent passes through (a, f(a)) with slope f′(a)
Height and slope come from different rulesIgnoring an open endpoint
✕Every algebraic root is an admissible input✓Filter roots against the stated domain, including endpoint brackets
The original function restriction still appliesDifficulty levels
introEvaluate a supplied or easily derived polynomial derivative
standardInterpret a bounded rate model and calculate an exact tangent
challengeFilter one, multiple or no gradient solutions and compare two kinds of rate
examHandle stationary multiplicity or a derivative-to-point-to-tangent chain with interpretation
Questions about this topic
Can there be no input with the required gradient?+
Yes. Algebraic roots may all be excluded by the stated domain
Does zero derivative always give a turning point?+
No. Check how the derivative sign behaves on both sides before drawing that conclusion
Are marks on a chain independent mastery evidence?+
No. Later parts depend on earlier calculations and may supply hints. They record practice activity until separately validated
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 12: Rates and derivatives
In this chapter
Calculate and interpret an average rate of change12A Average and instantaneous rates: Relate graph direction and instantaneous rateEstimate instantaneous rates with secants and tangentsDerive polynomial gradients from first principlesDifferentiate polynomialsApply polynomial derivatives12C Polynomial differentiation: Relate graph direction and instantaneous rate