E2.12 Differentiation

Learning objectives

E2.12.1Estimate gradients using tangents• Estimate gradients of curves by drawing tangents.E2.12.2Differentiate simple power functions• Use derivatives of functions of the form axⁿ, where a is rational and n is a non-negative integer, and simple sums of no more than three such terms; dy/dx notation is required.E2.12.3Gradients and stationary points• Apply differentiation to gradients and stationary points or turning points.E2.12.4Distinguish maxima and minima• Distinguish maxima and minima using an accurate sketch, the second derivative or the gradient on either side of a turning point; points of inflection are not required.

The gradient of a curve at one point is the gradient of its tangent there. A tangent follows the curve’s local direction but is treated as a straight line for the gradient calculation.

Mark the required point; place a ruler so the line just touches and matches the curve’s local direction; draw a long tangent; choose two well-separated, readable points on the tangent; calculate vertical change divided by horizontal change; include appropriate units.

$\text{gradient}=\frac{y_2-y_1}{x_2-x_1}$

The calculation points need not lie on the original curve. A longer triangle reduces the effect of reading error; retain the sign of the rise and run.

A rising tangent has positive gradient, a falling tangent negative gradient, and a horizontal tangent gradient $0$ . The estimate depends on the accuracy of both tangent and coordinate readings.

A chord through two curve points estimates an average gradient over an interval, not the instantaneous gradient at the named point.

The derivative $\dfrac{dy}{dx}$ gives the gradient function. For each allowed power term, multiply by the old power and reduce that power by one.

$\frac{d}{dx}(ax^n)=anx^{n-1}$

Original term	Derivative
$ax^n$	$anx^{n-1}$
$bx$	$b$
constant $c$	$0$

$y=5+8x-\frac43x^3\quad\Longrightarrow\quad\frac{dy}{dx}=8-4x^2$

Here $a$ is rational, $n$ is a non-negative integer, and the syllabus uses simple sums of no more than three such terms. Write the derivative in $dy/dx$ notation when required.

If $\frac{d}{dx}(3x^q)=15x^4$ , compare coefficient and power: $q-1=4$ and $3q=15$ , so $q=5$ .

Do not leave a constant in the derivative or reduce the coefficient instead of the power. The rule applies term by term to the stated polynomial scope.

Once a derivative is known, substituting an $x$ -value gives a gradient. A stationary point occurs where that gradient is zero.

Required result	Derivative move
gradient at $x=a$	calculate $dy/dx$ at $a$
points with gradient $m$	solve $dy/dx=m$ , then find each $y$
stationary points	solve $dy/dx=0$ , then substitute each $x$ into the original function
tangent equation	use derivative gradient and point in $y-y_1=m(x-x_1)$

$y=x^3-3x+4,\quad\frac{dy}{dx}=3x^2-3=0\quad\Longrightarrow\quad x=\pm1$

Substitute into the original function, not the derivative: $y(1)=2$ and $y(-1)=6$ , so the stationary points are $(1,2)$ and $(-1,6)$ .

If the curve passes through $(2,6)$ and the derivative gives gradient $7$ , then $y-6=7(x-2)$ , so the tangent is $y=7x-8$ .

Solving $dy/dx=0$ gives only the $x$ -coordinates. A complete stationary point requires the corresponding $y$ from the original function.

A stationary point is classified by how the function behaves around it. Cambridge accepts an accurate sketch, the gradient signs on either side, or the second derivative.

Evidence	Maximum	Minimum
gradient before and after	$+ o-$	$- o+$
second derivative at the point	$d^2y/dx^2<0$	$d^2y/dx^2>0$
accurate sketch	curve turns from rising to falling	curve turns from falling to rising

$\frac{dy}{dx}=3x^2-3,\qquad\frac{d^2y}{dx^2}=6x$

At $x=1$ , $d^2y/dx^2=6>0$ , so $(1,2)$ is a minimum. At $x=-1$ , $d^2y/dx^2=-6<0$ , so $(-1,6)$ is a maximum.

State both the classification and the evidence. Merely writing ‘maximum’ or ‘minimum’ does not show how the decision follows from the derivative or sketch.

Points of inflection are not required. If the second derivative is zero, this test alone is inconclusive; use an allowed alternative rather than assigning a maximum or minimum automatically.

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—E2.13 FunctionsRead

E2.12 Differentiation

Learning objectives

Estimate a curve gradient using a tangent

Differentiate simple power functions

Use derivatives for gradients, tangents and stationary points

Classify stationary points as maxima or minima

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