1.07a–1.07c - Derivatives as gradients and rates of change | OCR A-Level Mathematics A

Find a curve's gradient at one point by bringing two points together. You will connect the tangent and derivative in 1.07a–1.07b to instantaneous rates and their units, then sketch gradient functions in 1.07c without confusing height with slope.

From a chord to a tangent

How can a curve have a gradient at just one point, when calculating a gradient needs two points? On $y=x^2$ , the chord joining $(1,1)$ to $(2,4)$ has gradient $3$ . Move the second point to $(1.1,1.21)$ , and the gradient becomes $2.1$ . The two points are getting closer, and the slopes are settling towards $2$ .

You already know

A straight line through $(x_1,y_1)$ and $(x_2,y_2)$ , with $x_1\ne x_2$ , has gradient $(y_2-y_1)/(x_2-x_1)$ . A positive gradient rises as you move right; a negative gradient falls.

Keep one point fixed and let the other move along the curve. The straight lines joining them show how an average direction can approach a local direction.

Chords approaching the tangent to the parabola at A

Chord and tangent

A chord joins two distinct points of a curve. At a smooth point with a finite limiting chord gradient, the tangent is the straight line through that point with that gradient. A tangent need not meet the curve only once: the tangent to $y=x^3$ at $(1,1)$ also meets the curve at $(-2,-8)$ .

That last fact is a useful surprise. Tangency describes what happens nearby, not how many intersections occur elsewhere. For the parabola, the limit can be established exactly rather than guessed from rounded values.

Limiting chord gradient

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

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

Keep $h\ne0$ while simplifying. The first-principles limit definition is in the formula sheet.

In the formula sheet.

The limit asks what values approach; it never asks you to divide zero by zero.

Worked example 1: the tangent at one

Question. For $\mathrm f(x)=x^2$ , find the chord gradient between $A=(1,1)$ and $P=(1+h,(1+h)^2)$ . Hence find the tangent gradient at $A$ .

Plan. Form the coordinate differences, cancel $h$ , then take the limit.

\begin{aligned} m_{AP}&=\frac{(1+h)^2-1}{h} && \text{change in height over width}\\ &=\frac{2h+h^2}{h} && \text{expand and subtract}\\ &=2+h\quad(h\ne0) && \text{cancel a common factor}\\ \lim_{h\to0}m_{AP}&=2 && \text{approach from either side} \end{aligned}

Why this step? Cancelling while $h\ne0$ replaces the fraction by an expression whose limit is easy to see.

Answer. The chord gradient is $2+h$ ; the tangent gradient is $2$ .

Check. For $h=1,0.1,0.01$ , the slopes are $3,2.1,2.01$ ; for $h=-0.1,-0.01$ , they are $1.9,1.99$ . Both sides agree with the algebraic limit.

Change only the fixed point in the second example. The curve and the limiting argument stay the same.

Worked example 2: moving the fixed point

Question. Find the tangent gradient to $y=x^2$ at $A=(2,4)$ from a chord calculation.

Plan. Use the nearby point with horizontal coordinate $2+h$ .

\begin{aligned} m_{AP}&=\frac{(2+h)^2-4}{h} && \text{use the two curve points}\\ &=\frac{4h+h^2}{h}=4+h && \text{expand; cancel for }h\ne0\\ \lim_{h\to0}m_{AP}&=4 && \text{take the limit} \end{aligned}

Why this step? Both coordinates of the moving point must come from the same input $2+h$ .

Answer. The tangent gradient is $4$ .

Check. The chord slopes for $h=0.1$ and $h=-0.1$ are $4.1$ and $3.9$ , straddling $4$ .

This distinction between a nonzero increment and its limit prevents the most common first-principles error.

Watch out

Wrong: put $h=0$ into the unsimplified quotient. That gives $0/0$ , which is undefined. Right: simplify for $h\ne0$ , then find the value approached as $h\to0$ .

Try the same curve at a point on its falling side.

The derivative at a point and as a function

Instead of finding one tangent gradient, collect all of them into a function. For $\mathrm f(x)=x^2$ , the same chord algebra at a general input gives $2x+h$ ; taking its limit gives $2x$ .

Derivative

The derivative $\mathrm f'(x)$ is the function giving the tangent gradient wherever the finite derivative exists. If $y=\mathrm f(x)$ , write it equivalently as $\mathrm dy/\mathrm dx$ . The value $\mathrm f'(a)$ is the gradient at the particular input $a$ , whereas $\mathrm f(a)$ is the curve's height there.

Here $\mathrm f'(x)=2x$ follows from the limit, not from a rule you must assume. A derivative formula applies only where the function is defined and differentiable. At an endpoint, a one-sided rate may exist without a two-sided derivative.

Worked example 3: height and gradient

Question. Given $\mathrm f(x)=x^2$ and $\mathrm f'(x)=2x$ , state the point and tangent gradient when $x=3$ .

Plan. Substitute into the function for height and into its derivative for slope.

\begin{aligned} \mathrm f(3)&=3^2=9 && \text{height on the curve}\\ \mathrm f'(3)&=2(3)=6 && \text{tangent gradient} \end{aligned}

Why this step? The two formulas answer different questions even though their input is the same.

Answer. The point is $(3,9)$ ; its tangent gradient is $6$ .

Check. The nearby chord has slope $6+h$ , tending to $6$ .

Change only the input to a negative number. Positive height does not force a positive slope.

Worked example 4: above the axis but falling

Question. For the same function, state the point and tangent gradient when $x=-2$ , and interpret the sign.

Plan. Evaluate both formulas before describing the curve's direction.

\begin{aligned} \mathrm f(-2)&=(-2)^2=4 && \text{height}\\ \mathrm f'(-2)&=2(-2)=-4 && \text{slope} \end{aligned}

Why this step? Falling or rising is decided by the derivative, not by whether the point is above the axis.

Answer. At $(-2,4)$ , the tangent gradient is $-4$ , so the curve is locally falling as $x$ increases.

Check. Moving from $x=-2$ to $-1.9$ lowers the height from $4$ to $3.61$ .

The symbols for differentiation are a compact name for this limiting process.

Watch out

Wrong: $\mathrm dy/\mathrm dx=y/x$ . The derivative is a limit of ratios of changes, not the ratio of the current coordinates. For $y=x^2$ at $x=3$ , $y/x=3$ but the derivative is $6$ .

You can interpret derivative information even when no complete function is given.

Rates of change and their units

A derivative acquires practical meaning when the axes measure something. Suppose a tank contains $V(t)=20+3t^2$ litres after $t$ minutes, for $0\le t\le5$ . Its volume at $t=2$ is $32$ litres, but its filling rate is a different quantity.

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1.07a–1.07c - 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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