This note covers the derivative as the gradient of a curve and as an instantaneous rate of change, differentiating sums of power terms, finding tangents and normals, and deciding where a function is increasing or decreasing. Calculus lets you measure how fast a quantity such as speed, cost or population is changing at one exact moment.
Key ideas
The gradient of a straight line is the same everywhere. A curve bends, so its gradient changes. The gradient of a curve at a point is the gradient of the tangent, the straight line that just touches the curve there. The derivative \(f'(x)\) (also written \(\frac{dy}{dx}\)) is a function that gives this gradient at every value of \(x\).
- Rate of change. If \(y\) depends on \(x\), then \(\frac{dy}{dx}\) is the instantaneous rate of change of \(y\) with respect to \(x\). Its units are the units of \(y\) divided by the units of \(x\). For example, if distance \(s\) is in metres and time \(t\) is in seconds, \(\frac{ds}{dt}\) is in m s\(^{-1}\).
- Power rule. If \(f(x)=ax^{n}\), then \(f'(x)=anx^{n-1}\), for any integer \(n\) (including negative integers). At HL the rule also holds for fractional \(n\). The derivative of a constant is 0. Differentiate a sum term by term.
- Rewrite first. Write reciprocals as powers before differentiating: \(\frac{1}{x^{2}}=x^{-2}\). At HL, roots too: \(\sqrt{x}=x^{1/2}\), \(\frac{3}{\sqrt{x}}=3x^{-1/2}\). If a term is a fraction with a single term on the bottom, split it into separate powers.
- Gradient at a point. Substitute the \(x\)-value into \(f'(x)\). Your GDC can also find the gradient of a graph at a given point using its numerical derivative feature.
- Increasing and decreasing. \(f\) is increasing where \(f'(x)\gt 0\) and decreasing where \(f'(x)\lt 0\). Where \(f'(x)=0\) the tangent is horizontal, which can be a maximum, a minimum or a flat point.
- Tangent. At the point \((a,f(a))\), the tangent has gradient \(m=f'(a)\) and equation \(y-f(a)=f'(a)(x-a)\).
- Normal. The normal is perpendicular to the tangent at the same point. Its gradient is \(-\frac{1}{f'(a)}\) (provided \(f'(a)\neq 0\)), and its equation is \(y-f(a)=-\frac{1}{f'(a)}(x-a)\).
The techniques for finding tangents and normals follow a fixed pattern. Find \(f(a)\) for the \(y\)-coordinate, find \(f'(a)\) for the gradient, then substitute into the line equation. Always check that the point lies on the line.
Finding where the gradient has a given value. To find where the gradient of a curve equals a number \(k\), solve \(f'(x)=k\). For example, for \(y=x^{2}-6x+1\), the gradient is 4 where \(2x-6=4\), so \(x=5\). A horizontal tangent is the special case \(k=0\). Parallel tangents have the same gradient, and a tangent parallel to the line \(y=3x+1\) has gradient 3.
Reading a rate in context. A positive rate means the quantity is increasing and a negative rate means it is decreasing. A rate of \(-3\) litres per minute for the volume of water in a tank means the tank is losing 3 litres every minute at that moment. Always give the rate with its units and a sentence about what it means. The size of the rate is not the same as the size of the quantity: a population can be large and growing slowly.
Sketching. On a graph, steep parts have a large positive or negative gradient. Where the curve rises from left to right, \(f'\) is positive. Where it falls, \(f'\) is negative. At a smooth peak or valley, \(f'\) is zero.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f'(x)\) or \(\dfrac{dy}{dx}\) | The derivative: gradient of the curve and instantaneous rate of change. |
| \(\dfrac{d}{dx}(ax^{n})=anx^{n-1}\) | Power rule for any integer power \(n\) (fractional \(n\) at HL). |
| \(\dfrac{d}{dx}(c)=0\) | The derivative of a constant is zero. |
| \(y-y_1=m(x-x_1)\) | Equation of a line through \((x_1,y_1)\) with gradient \(m\). |
| Tangent gradient \(m=f'(a)\) | Gradient of the curve at \(x=a\). |
| Normal gradient \(=-\dfrac{1}{f'(a)}\) | Perpendicular to the tangent: the product of the two gradients is \(-1\). |
| \(f'(x)\gt 0\) / \(f'(x)\lt 0\) | The function is increasing / decreasing. |
Worked examples
Example 1 Let \(f(x)=2x^{3}-5x^{2}+4\). (a) Find the equation of the tangent at \(x=2\). (b) Find the equation of the normal at \(x=2\). (c) Find the values of \(x\) for which \(f\) is increasing.
Differentiate: \(f'(x)=6x^{2}-10x\).
(a) \(f(2)=2(8)-5(4)+4=16-20+4=0\), so the point is \((2,0)\). The gradient is \(f'(2)=6(4)-10(2)=24-20=4\).
Tangent: \(y-0=4(x-2)\), so \(y=4x-8\).
(b) The normal gradient is \(-\frac14\). Normal: \(y-0=-\frac14(x-2)\), so \(y=-\frac14x+\frac12\).
(c) We need \(f'(x)\gt 0\): \(6x^{2}-10x=2x(3x-5)\gt 0\). The roots of \(f'(x)=0\) are \(x=0\) and \(x=\frac53\). The quadratic \(2x(3x-5)\) opens upwards, so it is positive outside the roots.
Answer: (a) \(y=4x-8\) (b) \(y=-0.25x+0.5\) (c) \(x\lt 0\) or \(x\gt\frac53\)
Example 2 A fish farm's population is modelled by \(P(t)=1200+60t-2t^{2}\), where \(t\) is the time in months and \(t\ge 0\). (a) Find the rate of change of the population after 5 months. (b) Find when the population stops increasing, and the population then.
\(P'(t)=60-4t\), in fish per month.
(a) \(P'(5)=60-20=40\). The population is increasing at 40 fish per month at \(t=5\).
(b) The population stops increasing when \(P'(t)=0\): \(60-4t=0\), so \(t=15\). For \(t\lt 15\), \(P'(t)\gt 0\) so \(P\) increases. For \(t\gt 15\), \(P'(t)\lt 0\) so \(P\) decreases.
\(P(15)=1200+60(15)-2(225)=1200+900-450=1650\).
Answer: (a) 40 fish per month (b) at 15 months, when the population is 1650
Example 3 (HL, because of the fractional power) Find the gradient of \(g(x)=4\sqrt{x}+\dfrac{3}{x^{2}}\) at \(x=4\).
Rewrite as powers: \(g(x)=4x^{1/2}+3x^{-2}\).
Differentiate: \(g'(x)=4\times\frac12x^{-1/2}+3\times(-2)x^{-3}=2x^{-1/2}-6x^{-3}\).
At \(x=4\): \(x^{-1/2}=\frac{1}{\sqrt{4}}=\frac12\) and \(x^{-3}=\frac{1}{64}\).
\(g'(4)=2\times\frac12-6\times\frac{1}{64}=1-0.09375=0.90625\).
Answer: gradient \(\approx 0.906\) (exactly \(\frac{29}{32}\))
Common errors
- Differentiating \(\frac{3}{x^{2}}\) without rewriting it as \(3x^{-2}\), or forgetting to subtract 1 from a negative power (\(-2\) becomes \(-3\), not \(-1\)).
- Using \(f(a)\) as the gradient. The gradient of the tangent is \(f'(a)\); \(f(a)\) is the \(y\)-coordinate.
- Using the tangent gradient for the normal, or forgetting both the negative sign and the reciprocal for the normal.
- Leaving out units of a rate of change, or giving a rate with the wrong time unit.
- Solving \(f'(x)=0\) and stopping, instead of checking the sign of \(f'(x)\) to decide where the function increases or decreases.
Exam tip: For tangent and normal questions, find the point first, then the gradient, then write the line using \(y-y_1=m(x-x_1)\). Rearranging to \(y=mx+c\) is not needed unless the question asks for that form.
Check yourself
- Find the gradient of \(y=3x^{2}-7x\) at \(x=2\).
Answer
\(\frac{dy}{dx}=6x-7\). At \(x=2\): \(12-7=5\). - Find the equation of the tangent to \(y=x^{3}\) at the point where \(x=1\).
Answer
\(y'=3x^{2}=3\) at \(x=1\), and \(y=1\). Tangent: \(y-1=3(x-1)\), so \(y=3x-2\). - The distance travelled by a cyclist is \(s=3t^{2}-t\) metres after \(t\) seconds. Find the speed at \(t=4\).
Answer
\(\frac{ds}{dt}=6t-1\). At \(t=4\): \(24-1=23\) m s\(^{-1}\). - For which value of \(x\) is the tangent to \(y=x^{2}-6x+1\) horizontal?
Answer
\(y'=2x-6=0\), so \(x=3\).