5.3 Differentiating polynomials

Finding a gradient with tables of chords, as in 5.1, works for any function but is slow and only gives an estimate. For the most common functions there is a far quicker way. The gradient functions of $x^2$, $x^3$ and $x^4$ follow a simple pattern, and that pattern extends to every whole-number power of $x$, including negative powers. Together with two rules for sums and multiples, it allows any polynomial to be differentiated exactly in a single line.

These rules are the starting point for all of differential calculus. Tangents and normals (5.4), increasing and decreasing functions (5.2), maximum and minimum points (5.8) and motion (5.9) all depend on being able to differentiate quickly and accurately.

The derivative of $x^n$

A pattern in the gradient functions

The chord method of 5.1, applied to the first few powers of $x$, gives these gradient functions.

In each case the old power becomes a multiplier and the new power is one less than the old one. For example, the gradient of $y = x^3$ at $x = 2$ is $3 \times 2^2 = 12$, and a table of chords gives values approaching $12$: the chord from $x = 2$ to $x = 2.001$ has gradient $\frac{2.001^3 - 8}{0.001} = 12.006001$.

The pattern for $x^2$ can be explained with a picture. Think of $x^2$ as the area of a square of side $x$. If the side grows by a small amount $h$, the area grows by two strips of area $xh$ and a tiny corner of area $h^2$.

The rate of change of the area is the change in area divided by the change in side:

As $h \to 0$, this tends to $2x$, so the derivative of $x^2$ is $2x$. The same idea with a cube of side $x$ gives three thin slabs of volume $x^2h$ (plus much smaller pieces), so the derivative of $x^3$ is $3x^2$.

The power rule

The pattern holds for every power of $x$ where the power is an integer. Combined with a constant multiplier $a$, this gives the power rule.

Finding a derivative is called differentiation. To differentiate $y = f(x)$ "with respect to $x$" means to find $\frac{dy}{dx}$, or $f'(x)$.

The rule $f(x) = x^n \Rightarrow f'(x) = nx^{n-1}$ is in the formula booklet, where it holds for any real power $n$; 5.6 extends it to fractional powers.

Linear functions and constants

Two special cases of the power rule match what is already known about straight lines.

• A linear term. $ax = ax^1$, so its derivative is $a \times 1 \times x^0 = a$. The line $y = ax$ has gradient $a$ everywhere.

• A constant. The graph of $y = c$ is a horizontal line, which has gradient $0$. This agrees with the power rule, since $c = cx^0$ gives $c \times 0 \times x^{-1} = 0$

Why a constant multiplier stays

Multiplying a function by a constant $a$ stretches its graph vertically by a factor of $a$ (2.11). Every rise is multiplied by $a$ while every run stays the same, so every gradient is multiplied by $a$. This is the constant multiple rule: the derivative of $af(x)$ is $af'(x)$.

Differentiating polynomials

The sum rule

A polynomial is a sum of terms of the form $ax^n$. The gradient of a sum of two functions is the sum of their gradients, because the rise of the sum over any run is the sum of the two rises. This is the sum rule, and it means a polynomial is differentiated one term at a time.

The sum rule and the constant multiple rule together say that differentiation is "linear": it can be done term by term, and constants can be taken outside. Products and quotients do not behave like this.

Expanding brackets first

The power rule applies only to single terms $ax^n$. A product of brackets must first be expanded into a sum of such terms.

Dividing first

A fraction whose denominator is a single power of $x$ can be split into separate terms, each of which is a power of $x$.

Negative integer powers

Rewriting fractions as powers

The power rule also holds for negative integer powers, using the exponent law $\frac{1}{x^n} = x^{-n}$ (1.5). To differentiate a term with $x$ in the denominator, first rewrite it as a power of $x$, then apply the rule. For example,

The result agrees with the graph of the reciprocal function: both branches slope downwards everywhere, and $-\frac{1}{x^2}$ is negative for every $x \neq 0$. Near the vertical asymptote the branches are very steep, and $-\frac{1}{x^2}$ is large and negative; far from the origin they flatten out, and $-\frac{1}{x^2}$ tends to $0$.

The gradient at a point

Substituting into the derivative

The derivative $f'(x)$ is the gradient function, so the gradient of the curve at a particular point is found by substituting the $x$-coordinate of the point into $f'(x)$. Only the $x$-coordinate is used.

Points where the gradient has a given value

Solving $f'(x) = m$

To find where a curve has a particular gradient $m$, set the derivative equal to $m$ and solve for $x$. Then substitute each $x$-value into the original function to find the $y$-coordinates of the points. There may be no solutions, one, or several.

The two tangents in the graph are parallel, because they have the same gradient. "Find the points where the tangent is parallel to the line $y = 6x + 1$" is the same question as "find where the gradient is $6$". Finding the equations of these tangents is the subject of 5.4.

Negative powers and given gradients

Finding unknown coefficients

When a curve contains unknown constants, information about its gradient gives equations for them. A point on the curve gives an equation from $y = f(x)$, and a gradient gives an equation from $f'(x)$.

Rates of change with polynomial models

Exact rates of change

In a context, the derivative of a model gives the exact instantaneous rate of change, replacing the numerical estimates of 5.1. Two geometric examples show a pattern: the rate of change of the area of a circle with respect to its radius is its circumference, and the rate of change of the volume of a sphere with respect to its radius is its surface area.

Other variables and letters that stand for constants

The variable that a function is differentiated with respect to is the one in the denominator of the notation: $\frac{ds}{dt}$ is found by treating $s$ as a function of $t$. Any other letters in the formula, such as $a$, $k$, $g$ or $\pi$, are constants, and they are treated exactly like numbers.

Marginal cost

In economics, the derivative of a cost function $C(x)$ with respect to the number of items $x$ is called the marginal cost. It approximates the cost of producing one more item.

The variables in a model are often not $x$ and $y$, so the derivative is written with the letters of the context: $\frac{dA}{dr}$, $\frac{dC}{dx}$, $\frac{dh}{dt}$. The rules are exactly the same; only the names change.

Exam-style questions on differentiating polynomials

The second question uses a GDC to solve an equation in a context.

Self-check

• To do:

  I can state and use the power rule: if $f(x) = ax^n$, then $f'(x) = anx^{n-1}$ for $n \in \mathbb{Z}$.

• To do:

  I can explain why the derivative of a constant is $0$ and the derivative of $ax$ is $a$.

• To do:

  I can differentiate a polynomial term by term, expanding brackets or dividing first when needed.

• To do:

  I can rewrite terms such as $\frac{3}{x^2}$ and $\frac{1}{4x^3}$

• To do:

  I can find the gradient of a curve at a given point by substituting into the derivative.

• To do:

  I can find the points on a curve where the gradient takes a given value, including zero.

• To do:

  I can use information about a point and a gradient to find unknown coefficients.

• To do:

  I can interpret the derivative of a polynomial model as a rate of change with units.

1. (a)

   Find $f'(x)$.

   [2]2 marks

   Markscheme (a)

   $f'(x) = 3x^2 - 12x + 9$ A1A1

   Award A1 for each two correct terms.

2. (b)

   Find the coordinates of the two stationary points of the graph of $f$.

   [4]4 marks

   Markscheme (b)

   Setting $f'(x) = 0$: $3(x-1)(x-3) = 0$

3. (c)

   Find the equation of the tangent to the graph of $f$ at the point where $x = 4$.

   [3]3 marks

   Markscheme (c)

   $f'(4) = 48 - 48 + 9 = 9$

$4x^3$

(c) $g'(x) = -\frac{1}{2} \times 6x^5 = -3x^5$

(d) The variable is $r$, and $\frac{4}{3}\pi$ is a constant multiplier.

This is the surface area of a sphere of radius $r$, as estimated numerically in 5.1.

(b) Write $\frac{x^4}{8} = \frac{1}{8}x^4$ and $\frac{2x^3}{3} = \frac{2}{3}x^3$:

(c) $\frac{dy}{dx} = 0 - 1 - 5x^4 = -1 - 5x^4$.

(b) Expand: $f(x) = x^4 + 2x^2 + 1$, so $f'(x) = 4x^3 + 4x$.

(c) Expand: $(x - 2)^2 = x^2 - 4x + 4$, so $g(x) = x^3 - 4x^2 + 4x$ and $g'(x) = 3x^2 - 8x + 4$.

So $\frac{dy}{dx} = 2x - 4$.

(b) $y = \frac{2}{5}x^{-3}$, so $\frac{dy}{dx} = \frac{2}{5} \times (-3)x^{-4} = -\frac{6}{5}x^{-4} = -\dfrac{6}{5x^4}$.

(c) $y = 2x^2 - 3x^{-1} + x^{-4}$, so

The middle term: $-3 \times (-1)x^{-2} = 3x^{-2}$.

(d) Split the fraction: $g(x) = \frac{x^2}{x^2} - \frac{3}{x^2} = 1 - 3x^{-2}$. Then

$f'(-2) = 2 - \frac{8}{4} = 0$, so the tangent at $x = -2$ is horizontal: the graph is momentarily flat there. (The point $(-2, -8)$ is a stationary point; see 5.8.)

$f'(1) = 2 - 8 = -6$, so at $x = 1$ the graph is decreasing steeply, falling about $6$ units for each unit to the right.

Solve:

At $x = 2$, $y = 2 + \frac{1}{2} = \frac{5}{2}$; at $x = -2$, $y = -2 - \frac{1}{2} = -\frac{5}{2}$.

The points are $\left(2, \frac{5}{2}\right)$ and $\left(-2, -\frac{5}{2}\right)$.

The gradient is $\frac{dy}{dx} = 2ax + b$, and at $x = 1$ it equals $8$:

Subtracting the first equation from the second gives $a = 3$, and then $b = 2$.

Check: $y = 3x^2 + 2x$ gives $y = 5$ at $x = 1$ and $\frac{dy}{dx} = 6x + 2 = 8$ at $x = 1$.

(b) $k$ is a constant, so $kx^3$ differentiates to $3kx^2$, $-k^2x$ to $-k^2$, and $5k$ to $0$:

(c) Write $P = 3q^2 - 4q^{-1}$. Then

(c) $C(10) = 10 - 60 + 150 + 200 = 300$ and $C(11) = 13.31 - 72.6 + 165 + 200 = 305.71$. The 11th item costs $305.71 - 300 = \text{USD } 5.71$, close to the estimate of USD 6 from the derivative.

(d) Solve $0.03x^2 - 1.2x + 15 = 6$:

The marginal cost is also USD 6 when $x = 30$.

which is the given value (A1 for the working that leads to it).

(c) With $a = 3$, $f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x + 2)(x - 1)$.

$f'(x) = 0$ gives $x = 1$ (given) or $x = -2$.

$f(-2) = 2(-8) + 3(4) - 12(-2) + 1 = -16 + 12 + 24 + 1 = 21$. The point is $(-2, 21)$.

(d) Solve $6x^2 + 6x - 12 = -12$:

$f(0) = 1$ and $f(-1) = -2 + 3 + 12 + 1 = 14$. The points are $(0, 1)$ and $(-1, 14)$.

(c) Solve $d'(t) = 0.5$, that is, $-0.06t^2 + 0.6t - 0.9 = 0.5$. With the GDC (graphing $y = d'(t)$ and $y = 0.5$ and finding the intersections, or with the polynomial solver for $-0.06t^2 + 0.6t - 1.4 = 0$):

so $t = 3.71$ or $t = 6.29$ months (3 s.f.). Both are in the domain $0 \le t \le 12$.

$x = 1$ or $x = 3$ A1

$(1,\, 5)$ and $(3,\, 1)$ A1A1

$f(4) = 64 - 96 + 36 + 1 = 5$ A1

$y - 5 = 9(x - 4)$, so $y = 9x - 31$ A1

[2]2 marks

• y = h(x)