Mc1–Mc4 - Derivatives as gradients and rates of change | OCR A-Level Mathematics B (MEI)

Turn gradients between pairs of points into the gradient at one point. Mc1–Mc3 connect that tangent direction to the derivative, its notation and rates with units; Mc4 develops the gradient graph from the shape of a curve, including cases where flattening does not reverse its direction.

From a chord to a tangent

Can you measure a curve's slope at one point without pretending the curve is straight? On $y=x^2$ , the chord from $(1,1)$ to $(2,4)$ has gradient $3$ , but the chord to $(1.1,1.21)$ has gradient $2.1$ . Bringing the second point towards the first reveals a local direction.

You already know

For two distinct horizontal coordinates, a straight line's gradient is change in height divided by change in horizontal coordinate. A positive gradient rises to the right, a negative one falls, and a horizontal line has gradient zero.

Use a moving point on the curve, not a point chosen freely in the plane. Its height must change with its horizontal coordinate.

Chords approaching the tangent at a fixed parabola point

Chord and tangent

A chord joins two distinct curve points. A tangent, at a smooth point where a finite limiting chord gradient exists, is the line through the point with that gradient. It describes the local direction; it need not meet the curve only once. For example, the tangent to $y=x^3$ at $(1,1)$ meets the curve again at $(-2,-8)$ .

There is another useful surprise: lifting the entire graph by $1$ changes every height but none of its gradients. The two heights in a chord calculation both gain $1$ , so their difference is unchanged. That observation will help separate height from slope.

Tangent as a chord-gradient limit

For $A=(a,\mathrm f(a))$ , when the same finite limit exists from both sides,

\mathrm f'(a)=\lim_{h\to0}\frac{\mathrm f(a+h)-\mathrm f(a)}{h}.

Simplify with $h\ne0$ , then take the limit. The first-principles limit definition is supplied.

In the formula sheet.

A table suggests a value, but exact simplification explains why that value is approached.

Worked example 1: nearby points on a parabola

Question. Determine the tangent gradient to $y=x^2$ at $(1,1)$ by considering the point $(1+h,(1+h)^2)$ .

Plan. Express the chord gradient in terms of $h$ , then let the increment tend to zero.

\begin{aligned} m&=\frac{(1+h)^2-1}{h} && \text{matching coordinate changes}\\ &=\frac{2h+h^2}{h}=2+h && \text{expand; cancel for }h\ne0\\ \lim_{h\to0}m&=2 && \text{the same from both sides} \end{aligned}

Why this step? Taking the limit of the simplified expression avoids the undefined quotient obtained by substituting zero too early.

Answer. The tangent gradient is $2$ .

Check. The chord gradients for $h=1,0.1,0.01$ are $3,2.1,2.01$ , and for $h=-0.1,-0.01$ are $1.9,1.99$ .

Now change only the graph's vertical position. The chord calculation tests whether a higher graph must be steeper.

Worked example 2: the same slope at a different height

Question. Determine the tangent gradient to $y=x^2+1$ at $(1,2)$ using a nearby point.

Plan. Use $(1+h,(1+h)^2+1)$ and subtract the whole fixed height.

\begin{aligned} m&=\frac{(1+h)^2+1-2}{h} && \text{both points lie on the lifted curve}\\ &=\frac{2h+h^2}{h}=2+h && \text{the height shift cancels}\\ \lim_{h\to0}m&=2 && \text{take the limit} \end{aligned}

Why this step? A translation adds equally to both heights, leaving their difference unchanged.

Answer. The tangent gradient is still $2$ , although the point's height is now $2$ .

Check. At $h=0.1$ , the chord gradient is $(2.21-2)/0.1=2.1$ , as before.

The limit always concerns nonzero intervals. We are not replacing the chord by two identical points and then calculating a gradient.

Watch out

Wrong: substitute $h=0$ into the unsimplified difference quotient. Right: simplify while $h\ne0$ , then take the limit. A zero denominator is not a very small denominator.

Keep the lifted graph and change its fixed input.

The derivative at a point and as a function

A derivative packages all the available tangent gradients into one function. At a general input, the chord gradient for $y=x^2$ is $2x+h$ ; its limit is $2x$ . The lifted curve $y=x^2+1$ has the same gradient function.

Derivative

For $y=\mathrm f(x)$ , the derivative is written $\mathrm f'(x)$ or $\mathrm dy/\mathrm dx$ . It gives the tangent gradient at every input where that finite derivative exists. The value $\mathrm f(a)$ is a height; $\mathrm f'(a)$ is a gradient. For example, $\mathrm f(3)=9$ and $\mathrm f'(3)=6$ for $\mathrm f(x)=x^2$ .

A formula alone does not remove domain restrictions. Only evaluate a derivative where the function and derivative exist. At a domain endpoint, a one-sided rate may exist without a two-sided derivative.

Worked example 3: interpreting two outputs

Question. Given $\mathrm g(x)=x^2+1$ and $\mathrm g'(x)=2x$ , interpret $\mathrm g(2)$ and $\mathrm g'(2)$ geometrically.

Plan. Evaluate the height and gradient separately.

\begin{aligned} \mathrm g(2)&=2^2+1=5 && \text{curve height}\\ \mathrm g'(2)&=2(2)=4 && \text{tangent gradient} \end{aligned}

Why this step? The prime changes what the output represents, not the input being used.

Answer. The graph passes through $(2,5)$ , with tangent gradient $4$ , rising locally as $x$ increases.

Check. Its nearby chord gradient is $4+h$ , tending to $4$ .

Change only the sign of the input. The height remains positive while the direction reverses.

Worked example 4: a negative gradient above the axis

Question. Interpret $\mathrm g(-2)$ and $\mathrm g'(-2)$ for the same function.

Plan. Substitute carefully, keeping brackets around the negative input.

\begin{aligned} \mathrm g(-2)&=(-2)^2+1=5 && \text{height stays positive}\\ \mathrm g'(-2)&=2(-2)=-4 && \text{gradient changes sign} \end{aligned}

Why this step? The sign of the gradient describes direction; the sign of the height describes position relative to the axis.

Answer. The graph passes through $(-2,5)$ and is locally falling to the right with tangent gradient $-4$ .

Check. At $x=-1.9$ , its height is $4.61$ , lower than $5$ .

You may also meet a derivative value in a table without knowing a formula. It has exactly the same meaning.

Watch out

Wrong: read $\mathrm dy/\mathrm dx$ as $y/x$ . Right: it denotes a limit of ratios of changes. On $y=x^2$ , at $x=3$ , the coordinate ratio is $3$ , while the derivative is $6$ .

Interpret the following information as two separate graph facts.

Rates of change and their units

Suppose a cooling model gives $T(t)=30-t^2$ , where $T$ is in degrees Celsius and $t$ is in minutes, for $0\le t\le4$ . Between two times, the temperature difference gives an average rate. At a single interior time, the derivative gives the instantaneous cooling rate.

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Mc1–Mc4 - Derivatives as gradients and rates of change

From a chord to a tangent

The derivative at a point and as a function

Rates of change and their units

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