This note covers differentiation of expressions made from powers of \(x\), finding gradients, tangents and normals, locating and classifying stationary points, and using these ideas in optimisation and motion problems. Differentiation measures how fast one quantity changes with another, which is central to physics, economics and engineering.
Key ideas
The derivative of \(y\) with respect to \(x\), written \(\dfrac{dy}{dx}\) (or \(f'(x)\) when \(y = f(x)\)), gives the gradient of the curve at any point. It tells us how quickly \(y\) changes as \(x\) changes.
The power rule. If \(y = ax^n\), then \[ \frac{dy}{dx} = anx^{n-1} \] Multiply by the power, then reduce the power by 1. This works for positive, negative and fractional powers. Other points:
- A constant differentiates to 0, and \(x\) differentiates to 1.
- Differentiate an expression term by term.
- Rewrite roots and fractions as powers first: \(\sqrt{x} = x^{1/2}\) and \(\dfrac{1}{x^3} = x^{-3}\).
- Expand brackets or split fractions before differentiating. For example \(\dfrac{x^2+4}{x} = x + 4x^{-1}\).
Gradient at a point. Differentiate, then substitute the \(x\)-value into \(\dfrac{dy}{dx}\).
Tangent and normal. The tangent to a curve at a point has the same gradient as the curve there. The normal is the line through that point perpendicular to the tangent, so its gradient is \(-\dfrac{1}{m}\), where \(m\) is the gradient of the tangent. Use \(y - y_1 = m(x - x_1)\) for both lines.
Stationary points. At a stationary point the gradient is zero: \(\dfrac{dy}{dx} = 0\). Solve this equation to find the \(x\)-values, then substitute into the original equation to find the \(y\)-values. There are three types: maximum, minimum and point of inflection.
Finding the nature of a stationary point.
- Second derivative test. Differentiate again to get \(\dfrac{d^2y}{dx^2}\). If it is positive at the point, it is a minimum. If it is negative, it is a maximum. If it is zero, the test does not decide, so check the gradient on each side instead.
- Gradient test. Find the sign of \(\dfrac{dy}{dx}\) just before and just after the point. Positive then negative is a maximum. Negative then positive is a minimum.
Optimisation. To find the greatest or least value of a quantity, write it as a function of one variable, differentiate, set the derivative to zero and solve. Then confirm whether the result is a maximum or minimum, and answer the question in context.
Rates of change in motion. If \(s\) is the distance of an object from a fixed point at time \(t\), then the velocity is \(v = \dfrac{ds}{dt}\) and the acceleration is \(a = \dfrac{dv}{dt}\). Differentiating once gives velocity, and twice gives acceleration.
Increasing and decreasing. A function is increasing where \(\dfrac{dy}{dx} \gt 0\) and decreasing where \(\dfrac{dy}{dx} \lt 0\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\dfrac{d}{dx}(ax^n) = anx^{n-1}\) | Differentiating a power of \(x\). |
| \(\dfrac{dy}{dx}\) or \(f'(x)\) | Gradient function: the gradient of the curve at each point. |
| \(\dfrac{d^2y}{dx^2}\) or \(f''(x)\) | Second derivative: the derivative of the derivative. |
| Stationary point | A point where \(\dfrac{dy}{dx} = 0\). |
| \(\dfrac{d^2y}{dx^2} \gt 0\) | Minimum point. |
| \(\dfrac{d^2y}{dx^2} \lt 0\) | Maximum point. |
| Normal gradient \(= -\dfrac{1}{m}\) | Used when the tangent has gradient \(m\). |
| \(v = \dfrac{ds}{dt},\ a = \dfrac{dv}{dt}\) | Velocity and acceleration from displacement. |
Worked examples
Example 1 A curve has equation \(y = 2x^3 - 9x^2 + 12x + 1\). Find the coordinates of its stationary points and determine their nature.
Differentiate: \(\dfrac{dy}{dx} = 6x^2 - 18x + 12\).
Set to zero: \(6x^2 - 18x + 12 = 0\). Divide by 6: \(x^2 - 3x + 2 = 0\), so \((x-1)(x-2) = 0\).
So \(x = 1\) or \(x = 2\).
When \(x = 1\): \(y = 2 - 9 + 12 + 1 = 6\). When \(x = 2\): \(y = 16 - 36 + 24 + 1 = 5\).
Second derivative: \(\dfrac{d^2y}{dx^2} = 12x - 18\).
At \(x = 1\): \(12 - 18 = -6\), which is negative, so a maximum.
At \(x = 2\): \(24 - 18 = 6\), which is positive, so a minimum.
Answer: maximum at \((1, 6)\); minimum at \((2, 5)\)
Example 2 The curve \(y = x^2 + \dfrac{8}{x}\) passes through the point \(P\) where \(x = 2\). Find the equations of the tangent and the normal at \(P\).
Rewrite: \(y = x^2 + 8x^{-1}\).
At \(x = 2\): \(y = 4 + 4 = 8\), so \(P = (2, 8)\).
Differentiate: \(\dfrac{dy}{dx} = 2x - 8x^{-2} = 2x - \dfrac{8}{x^2}\).
At \(x = 2\): \(\dfrac{dy}{dx} = 4 - \dfrac{8}{4} = 2\).
Tangent: \(y - 8 = 2(x - 2)\), so \(y = 2x + 4\).
The normal gradient is \(-\tfrac12\). Normal: \(y - 8 = -\tfrac12(x - 2)\), so \(y = -\tfrac12 x + 9\).
Check: at \(x=2\) the normal gives \(-1 + 9 = 8\). Correct.
Answer: tangent \(y = 2x + 4\); normal \(y = -\tfrac12 x + 9\)
Example 3 A farmer has 60 m of fencing and uses a straight wall as one side of a rectangular pen. The two sides at right angles to the wall each have length \(x\) metres. Find the value of \(x\) that gives the largest area, and that area.
The side parallel to the wall uses the rest of the fencing: \(60 - 2x\).
Area: \(A = x(60 - 2x) = 60x - 2x^2\).
Differentiate: \(\dfrac{dA}{dx} = 60 - 4x\).
Set to zero: \(60 - 4x = 0\), so \(x = 15\).
Second derivative: \(\dfrac{d^2A}{dx^2} = -4\), which is negative, so this is a maximum.
The side parallel to the wall is \(60 - 30 = 30\) m. Area \(= 15 \times 30 = 450\) m².
Answer: \(x = 15\) m, giving a maximum area of 450 m²
Common errors
- Common error: forgetting to rewrite \(\dfrac{1}{x^2}\) or \(\sqrt{x}\) as powers before differentiating.
- Common error: differentiating \(\dfrac{x^2+4}{x}\) by differentiating the top and bottom separately. Split it into \(x + 4x^{-1}\) first.
- Common error: finding the \(x\)-value of a stationary point and stopping. If the question asks for coordinates, substitute into the original equation, not into \(\dfrac{dy}{dx}\).
- Common error: mixing up the sign of the second derivative: positive means minimum, negative means maximum.
- Common error: using the tangent gradient for the normal, or forgetting to change the sign when flipping.
Exam tip: In an optimisation question, write the quantity as a function of one letter before you differentiate, and finish by answering in the context of the problem, with units.
Check yourself
- Differentiate \(y = 5x^4 - 3x^2 + 7x - 2\).
Answer
\(\dfrac{dy}{dx} = 20x^3 - 6x + 7\). - Find the gradient of \(y = 3\sqrt{x} + \dfrac{2}{x^2}\) at \(x = 4\).
Answer
\(y = 3x^{1/2} + 2x^{-2}\), so \(\dfrac{dy}{dx} = \tfrac32 x^{-1/2} - 4x^{-3}\). At \(x = 4\): \(\tfrac32 \times \tfrac12 - \tfrac{4}{64} = 0.75 - 0.0625 = 0.6875\), which is \(\tfrac{11}{16}\). - The displacement of a particle is \(s = t^3 - 6t^2 + 9t\) metres at \(t\) seconds. Find when it is at rest and its acceleration at \(t = 4\).
Answer
\(v = \dfrac{ds}{dt} = 3t^2 - 12t + 9 = 3(t-1)(t-3)\). At rest when \(v = 0\), so at \(t = 1\) and \(t = 3\) seconds. Acceleration \(a = 6t - 12\), so at \(t = 4\), \(a = 12\) m/s². - Find the stationary point of \(y = x^2 - 6x + 11\) and state its nature.
Answer
\(\dfrac{dy}{dx} = 2x - 6 = 0\), so \(x = 3\) and \(y = 9 - 18 + 11 = 2\). The second derivative is 2, which is positive, so \((3, 2)\) is a minimum.