1.7 - From chords to a tangent | Cambridge A-Level Mathematics

A curve does not have one gradient for its whole length. You will find its gradient at a particular point by bringing a second point closer, then connect this idea to first and second derivatives. This gives a precise meaning to the slope you see in a sketch.

A chord approaches a tangent

What is the gradient of $y=x^2$ exactly at $(1,1)$ ? A single point gives no rise and run. Start with a second point on the curve, calculate the gradient of the line joining them, then move that second point closer.

You already know

The gradient of a non-vertical straight line through two points is change in $y$ divided by change in $x$ . Use the same subtraction order in the numerator and denominator.

The line segment joining two points on a curve is a chord. Keep $P=(1,1)$ fixed and let $Q=(1+h,(1+h)^2)$ , where $h\ne0$ . Its gradient is

\frac{(1+h)^2-1}{h}=2+h.

Here $h$ is a horizontal change, not a point's coordinate. Positive $h$ puts $Q$ to the right; negative $h$ puts it to the left.

[DIAGRAM: asset_name: Chords approaching a tangent; asset_slug: chords-to-tangent; caption: Moving Q towards P brings the chord gradient towards 2 from either side; description: Two plots of y equals x squared near P at 1 comma 1. The left shows chords from P to points with x coordinates 2 and 1.5, with gradients 3 and 2.5. The right shows points with x coordinates 0.5 and 0.9, with gradients 1.5 and 1.9. Both show the limiting tangent y equals 2x minus 1 with gradient 2.]
Moving Q towards P brings the chord gradient towards 2 from either side

Worked example 1: approaching from the right

Question. For $y=x^2$ , find the gradients of chords from $P=(1,1)$ to points with $x=2$ , $1.5$ and $1.1$ .

Plan. Calculate each rise over its horizontal change from $P$ .

\begin{aligned} m_1&=\frac{4-1}{2-1}=3\\ m_2&=\frac{2.25-1}{1.5-1}=2.5\\ m_3&=\frac{1.21-1}{1.1-1}=2.1 \end{aligned}

Answer. The gradients are $3$ , $2.5$ and $2.1$ , approaching $2$ .

Check. The horizontal changes are $1$ , $0.5$ and $0.1$ . Adding each to $2$ gives the same gradients.

Now change only the side from which $Q$ approaches. Both rise and run will be negative.

Worked example 2: approaching from the left

Question. For the same curve and fixed point, find chord gradients when $Q$ has $x=0.5$ , $0.9$ and $0.99$ .

Plan. Keep the order $Q-P$ in both differences.

\begin{aligned} m_1&=\frac{0.25-1}{0.5-1}=1.5\\ m_2&=\frac{0.81-1}{0.9-1}=1.9\\ m_3&=\frac{0.9801-1}{0.99-1}=1.99 \end{aligned}

Answer. The gradients approach $2$ from below.

Check. With $h=-0.5,-0.1,-0.01$ , the expression $2+h$ agrees with all three results. Two negative differences have given a positive gradient.

The same limiting value appears from both sides. This is stronger evidence than one isolated chord.

Watch out

At $Q=P$ , the chord calculation becomes $0/0$ , which is undefined. Bring distinct points closer; do not call $0/0$ the gradient.

Try a different fixed point on the same curve.

The derivative at a point

The value approached by the chord gradients is called a limit. You only need the informal idea here: as the horizontal gap becomes arbitrarily small, the gradients settle towards one value. The resulting line through the fixed point is the tangent.

Derivative at a point

The derivative $f'(a)$ is the gradient of the tangent to $y=f(x)$ at $x=a$ , when that finite limiting gradient exists. It is approached by $[f(a+h)-f(a)]/h$ as nonzero $h$ approaches zero.

A tangent describes the curve's local direction. It need not meet the whole curve only once, and it may cross a curve. Its defining feature is its limiting direction at the chosen point.

For $y=x^3$ at $P=(2,8)$ , expand the specific chord calculation:

\begin{aligned} \frac{(2+h)^3-8}{h} &=\frac{12h+6h^2+h^3}{h}\\ &=12+6h+h^2\qquad(h\ne0). \end{aligned}

As $h$ approaches zero, the last two terms approach zero. The chord gradient approaches $12$ . We have justified this particular limiting value without needing a general first-principles differentiation procedure.

Worked example 3: a cubic from the right

Question. Use $h=0.1$ and $h=0.01$ to investigate the tangent gradient of $y=x^3$ at $x=2$ .

Plan. Substitute the two gaps into the simplified chord gradient.

\begin{aligned} h=0.1 &: \quad 12+0.6+0.01=12.61\\ h=0.01 &: \quad 12+0.06+0.0001=12.0601 \end{aligned}

Answer. These gradients approach $12$ , so they support $f'(2)=12$ .

Check. Directly, $(2.1^3-8)/0.1=(9.261-8)/0.1=12.61$ .

Change only the sign of the gaps to check the other side.

Worked example 4: a cubic from the left

Question. Repeat the investigation with $h=-0.1$ and $h=-0.01$ .

Plan. Substitute the negative gaps before evaluating their squares.

\begin{aligned} h=-0.1 &: \quad 12-0.6+0.01=11.41\\ h=-0.01 &: \quad 12-0.06+0.0001=11.9401 \end{aligned}

Answer. The same limit $12$ is approached from below.

Check. Directly, $(1.9^3-8)/(-0.1)=(-1.141)/(-0.1)=11.41$ . The sign is positive, as the curve rises near $P$ .

The limiting expression explains the value; rounding one nearby chord does not establish an exact derivative. At a sharp corner, for example, the two sides can give different limits, so a derivative need not exist at every point of every curve.

Key idea

$f(2)=8$ is a height; $f'(2)=12$ is a gradient. They answer different questions about the same point.

Use the difference between a nearby chord and its limiting value.

First and second derivatives

Repeat the tangent calculation at different inputs and you obtain a new function: the gradient function. For $f(x)=x^2$ , the tangent gradients at $x=-1,0,1$ are $-2,0,2$ . The general gradient function is $f'(x)=2x$ , which you will learn to calculate directly next.

First and second derivatives

For $y=f(x)$ , the first derivative is written $f'(x)$ or $\dfrac{dy}{dx}$ . Differentiating the gradient function gives the second derivative, written $f''(x)$ or $\dfrac{d^2y}{dx^2}$ . It measures how the gradient changes as $x$ changes.

The superscript $2$ in $d^2y/dx^2$ records differentiating twice. It does not mean squaring $dy/dx$ . In the quadratic example, $f'(x)=2x$ is a straight line of gradient $2$ , so $f''(x)=2$ .

[DIAGRAM: asset_name: A curve and its gradient function; asset_slug: first-second-derivatives; caption: Tangent slopes on the parabola become heights on its gradient graph; description: Left plot shows y equals x squared, tangents at minus 1 comma 1 and 1 comma 1, and labelled points minus 2 comma 3 and 0 comma minus 1 on the first tangent and 0 comma minus 1 and 2 comma 3 on the second. Right plot shows f prime of x equals 2x through minus 1 comma minus 2, 0 comma 0 and 1 comma 2, with a rise of 4 over a run of 2.]
Tangent slopes on the parabola become heights on its gradient graph

When estimating a gradient from a drawn curve, draw its tangent at the required point, then choose two well-separated readable points on that tangent. They need not lie on the curve.

Worked example 5: estimating a positive derivative

Question. A tangent to $y=f(x)$ at $x=1$ passes through $(0,-1)$ and $(2,3)$ . Estimate $f'(1)$ .

Plan. Find the gradient of the tangent using its two stated points.

f'(1)\approx\frac{3-(-1)}{2-0}=2.

Answer. $f'(1)\approx2$ .

Check. The tangent rises $4$ units over $2$ units horizontally, so a positive gradient of $2$ fits the drawing.

Change only the contact point to the left side of the parabola. The tangent now falls as you move right.

Worked example 6: estimating a negative derivative

Question. A tangent at $x=-1$ passes through $(-2,3)$ and $(0,-1)$ . Estimate $f'(-1)$ , then use $f'(-1)=-2$ and $f'(1)=2$ to find the gradient of the straight graph $y=f'(x)$ .

Plan. Find the tangent slope first, then the slope between the two points on the gradient graph.

\begin{aligned} f'(-1)&\approx\frac{-1-3}{0-(-2)}=-2\\ \text{gradient of }y=f'(x)&=\frac{2-(-2)}{1-(-1)}=2. \end{aligned}

Answer. $f'(-1)\approx-2$ ; the second derivative is $2$ for this straight gradient graph.

Check. Moving $2$ units along the gradient graph increases its height by $4$ , agreeing with $f''(x)=2$ .

The opening question now has an answer: the gradient at $(1,1)$ on $y=x^2$ is $2$ , although the curve's gradient changes elsewhere.

Watch out

Use points on the tangent to estimate $f'(a)$ . Joining two distant points on the curve gives a chord gradient instead. A graph estimate should be reported with suitable precision, not as an unjustified exact value.

Check both the notation and the slope calculation.

Quick Recap

Chord gradients use two distinct points on the curve.
The tangent gradient is their limiting value as the second point approaches the first.
$f'(a)$ and $dy/dx$ describe a first derivative; $f''(a)$ and $d^2y/dx^2$ describe a second derivative.
Estimate a curve's gradient using two readable points on its tangent.

At $(1,1)$ on $y=x^2$ , the chord gradients approach $2$ .

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1.7 - From chords to a tangent

A chord approaches a tangent

The derivative at a point

First and second derivatives

Quick Recap

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