Differentiation – OCR FSMQ Additional Maths

1 The Gradient of a Curve

A straight line has one gradient everywhere. A curve's steepness changes from point to point, so "the gradient of the curve" only makes sense at a particular point — and it means the gradient of the tangent there (CA3).

CA2 — what the derivative means

$\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is the gradient function: substitute an $x$-value and it gives the gradient there.
It also measures the rate of change of $y$ with respect to $x$.

The two readings matter equally. Geometrically, the derivative is a gradient. In a context, it is a rate: if $V$ is volume and $t$ is time, $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ is how fast the volume is changing, in units of volume per unit time.

2 The Rule for $kx^n$

CA1 — the only differentiation rule you need

$y = kx^n \;\Longrightarrow\; \dfrac{\mathrm{d}y}{\mathrm{d}x} = knx^{n-1}$
Multiply by the power, then knock one off the power.

And for sums: differentiate each term separately and add the results. That is all you need, because the specification restricts you to sums of terms of the form $kx^n$.

$y$	$\dfrac{\mathrm{d}y}{\mathrm{d}x}$	Why
$x^5$	$5x^4$	Power $5$ comes down, becomes $4$
$7x^3$	$21x^2$	$7 \times 3 = 21$
$x$	$1$	$x = x^1 \to 1x^0 = 1$
$6x$	$6$	The gradient of a straight line
$9$	$0$	A constant does not change
$-4x^2$	$-8x$	The sign is carried through

The three traps. $\dfrac{\mathrm{d}}{\mathrm{d}x}(9) = 0$, not $9$. $\dfrac{\mathrm{d}}{\mathrm{d}x}(6x) = 6$, not $6x$ or $0$. And a power of $1$ reduces to a power of $0$, which is $1$ — not zero.

Worked Example 1 — A polynomial

Differentiate $y = 5x^4 - 2x^3 + 7x - 11$.

①$5x^4 \to 5 \times 4 x^3 = 20x^3$

②$-2x^3 \to -2 \times 3 x^2 = -6x^2$

③$7x \to 7$

④$-11 \to 0$

⑤$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 20x^3 - 6x^2 + 7$

Degree check: differentiating a quartic must give a cubic. If your answer still has an $x^4$, you have slipped.

Worked Example 2 — Expand before differentiating

Differentiate $y = x^2(3x - 4)$.

①There is no product rule in this course, so expand first:

②$y = 3x^3 - 4x^2$

③$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 9x^2 - 8x$

You may not differentiate each bracket separately. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is not $2x \times 3$. Expanding is compulsory, and always possible for the functions in this specification.

Worked Example 3 — Using function notation

$\mathrm{f}(x) = 2x^3 - 9x^2 + 12x$. Find $\mathrm{f}'(x)$ and evaluate $\mathrm{f}'(1)$ and $\mathrm{f}'(2)$.

①$\mathrm{f}'(x) = 6x^2 - 18x + 12$

②$\mathrm{f}'(1) = 6 - 18 + 12 = 0$ — the curve is momentarily flat at $x = 1$.

③$\mathrm{f}'(2) = 24 - 36 + 12 = 0$ — flat again at $x = 2$.

④Both are stationary points. Between them, $\mathrm{f}'(1.5) = 13.5 - 27 + 12 = -1.5 < 0$, so the curve is falling there.

3 What the Sign of the Derivative Tells You

$\dfrac{\mathrm{d}y}{\mathrm{d}x}$	The curve is…
Positive	Increasing — going uphill as $x$ increases
Negative	Decreasing — going downhill
Zero	Stationary — momentarily flat
Large in size	Steep
Close to zero	Nearly flat

Worked Example 4 — Where is a curve increasing?

Find the values of $x$ for which $y = x^3 - 12x$ is increasing.

①Increasing means $\dfrac{\mathrm{d}y}{\mathrm{d}x} > 0$.

②$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12 > 0$

③$3\left(x^2 - 4\right) > 0$, so $(x-2)(x+2) > 0$.

④A positive quadratic is above zero outside its roots:

⑤$x < -2$ or $x > 2$

Notice this is an quadratic inequality question in disguise. The calculus provides the inequality; the algebra solves it.

Worked Example 5 — A rate of change in context

A balloon's volume after $t$ seconds is $V = 4t^2 + 3t$ cm³. Find the rate at which it is inflating at $t = 5$, and state the units.

①$\dfrac{\mathrm{d}V}{\mathrm{d}t} = 8t + 3$

②At $t = 5$: $40 + 3 = 43$

③The rate is $43$ cm³ per second.

Always state units for a rate. They are the units of the top variable divided by the units of the bottom one — here cm³ per second.

4 The Second Derivative

Differentiating twice

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ or $\mathrm{f}''(x)$ — the rate of change of the gradient

If the first derivative says whether the curve is going up or down, the second says whether it is bending upwards or downwards. That is what makes it the test for maxima and minima.

Worked Example 6 — Two derivatives

$y = x^4 - 2x^3 + 5x$. Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 - 6x^2 + 5$

②Differentiate again: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x^2 - 12x$

③Note the $+5$ vanished at the second step, as every constant does.

5 Quick Reference

The rule

$kx^n \to knx^{n-1}$.

Sums

Differentiate term by term.

$kx$

Differentiates to $k$.

Constants

Differentiate to $0$.

Brackets

Expand first — no product rule here.

Notation

$\dfrac{\mathrm{d}y}{\mathrm{d}x}$, $\mathrm{f}'(x)$, $\dot{x}$.

Gradient at a point

Substitute the $x$-value.

Positive derivative

Increasing.

Rates

State the units.

Degree check

Differentiating drops the degree by one.

6 Practice Questions

Question 1

Differentiate $y = x^7$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 7x^6$

Question 2

Differentiate $y = 4x^3 - 6x^2 + 2x - 9$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^2 - 12x + 2$

Question 3

Find $\mathrm{f}'(x)$ when $\mathrm{f}(x) = 5 - 3x + x^2$.

▶ Show solution

$\mathrm{f}'(x) = -3 + 2x$, or $2x - 3$.

Question 4

Differentiate $y = x(x+5)$.

▶ Show solution

Expand: $y = x^2 + 5x$

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x + 5$

Question 5

Find the gradient of $y = 2x^3 - x^2$ at $x = -1$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 2x$

At $x = -1$: $6(1) - 2(-1) = 6 + 2 = 8$

Question 6

Find $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ when $y = x^4 - 3x^2$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 - 6x$

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x^2 - 6$

Question 7

Find the value of $x$ at which $y = x^2 - 10x$ has gradient $4$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 10 = 4$

$2x = 14$, so $x = 7$.

Question 8

Find the values of $x$ for which $y = x^3 - 3x^2$ is decreasing.

▶ Show solution

Decreasing means $\dfrac{\mathrm{d}y}{\mathrm{d}x} < 0$.

$3x^2 - 6x < 0 \;\Rightarrow\; 3x(x-2) < 0$

Between the roots: $0 < x < 2$

Question 9

Differentiate $y = (2x-1)(x+3)$ and hence find the gradient at $x = 2$.

▶ Show solution

Expand: $y = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3$

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x + 5$

At $x = 2$: $8 + 5 = 13$

Question 10

A cylindrical tank is being filled. After $t$ minutes the depth of water is $h = 0.5t^2 - 0.02t^3$ cm, valid for $0 \leqslant t \leqslant 16$.

(a) Find $\dfrac{\mathrm{d}h}{\mathrm{d}t}$ and state what it represents, with units. (b) Find the rate of filling at $t = 5$ and at $t = 15$. (c) Find when the water level stops rising. (d) Explain what the model predicts after that time, and why the restriction $t \leqslant 16$ is sensible.

▶ Show solution

(a) $\dfrac{\mathrm{d}h}{\mathrm{d}t} = t - 0.06t^2$

This is the rate at which the depth is increasing, in cm per minute.

(b) At $t = 5$: $5 - 0.06(25) = 5 - 1.5 = 3.5$ cm min⁻¹

At $t = 15$: $15 - 0.06(225) = 15 - 13.5 = 1.5$ cm min⁻¹

The tank is filling more slowly at $t = 15$ than at $t = 5$.

(c) The level stops rising when the rate is zero:

$t - 0.06t^2 = 0 \;\Rightarrow\; t(1 - 0.06t) = 0$

$t = 0$ or $t = \dfrac{1}{0.06} = 16.67$ minutes.

$t = 0$ is the start (the tank begins at rest), so the level stops rising at $t = 16.7$ minutes (3 s.f.).

(d) Beyond $t = 16.67$, the derivative becomes negative, so the model predicts the depth would fall — water leaving the tank of its own accord.

That is physically impossible for a tank being filled, so the model has broken down. Restricting to $t \leqslant 16$ keeps the model inside the range where the rate is still positive, which is exactly the sort of limitation on a model that AO3 expects you to identify and comment on.

Differentiation (CA1–CA3) · OCR FSMQ Additional Maths · Created with MathJax