5.3 Differentiating polynomials (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

The power rule

For any integer $n$ and constant $a$ : $f(x)=ax^{n}\ \Rightarrow\ f'(x)=anx^{n-1}.$ Multiply by the power, then reduce the power by one.

$f(x)=5x^{4}\Rightarrow f'(x)=20x^{3}$
$f(x)=7x\Rightarrow f'(x)=7$ (since $x^{0}=1$ )
$f(x)=9$ (a constant) $\Rightarrow f'(x)=0$ : a horizontal line has gradient zero.

Key termspower rule

Common mistake

Reducing the power without multiplying: the derivative of $x^{3}$ is $3x^{2}$ , not $x^{2}$ .

Section 2

Sums and differences of terms

Differentiate a polynomial term by term: $\frac{\mathrm{d}}{\mathrm{d}x}\left(ax^{n}+bx^{n-1}+\cdots\right)=anx^{n-1}+b(n-1)x^{n-2}+\cdots$ Example: $f(x)=4x^{3}-5x^{2}+7\Rightarrow f'(x)=12x^{2}-10x$ .

The derivative of a sum is the sum of the derivatives, and a constant multiple stays in front.

Key termsterm by term

Exam tip

Substitute negative values in brackets: $f'(-1)=12(-1)^{2}-10(-1)=22$ .

Section 3

Negative powers

The rule works for negative integer powers. First rewrite fractions as powers of $x$ : $\frac{8}{x^{2}}=8x^{-2}\ \Rightarrow\ \frac{\mathrm{d}}{\mathrm{d}x}=8(-2)x^{-3}=-\frac{16}{x^{3}}.$ Reducing a negative power by one makes it more negative: $-2\to-3$ , not $-1$ .

Example: $g(x)=\frac{8}{x^{2}}-3x\Rightarrow g'(x)=-\frac{16}{x^{3}}-3$ .

Key termsnegative power

Common mistake

Writing the derivative of $x^{-2}$ as $-2x^{-1}$ . The power goes down by one: $-2x^{-3}$ .

Common mistake

Differentiating $\frac{1}{x^{2}}$ as $\frac{1}{2x}$ : always rewrite as $x^{-2}$ first.

Section 4

Products and quotients: simplify first

At this stage you can only differentiate sums of $ax^{n}$ terms, so expand or divide first.

Product: $(x^{2}-3)(2x+1)=2x^{3}+x^{2}-6x-3$ , so $\frac{\mathrm{d}y}{\mathrm{d}x}=6x^{2}+2x-6$ .
Quotient by a single term: $\frac{x^{3}+4x}{x^{2}}=x+4x^{-1}$ , so $\frac{\mathrm{d}y}{\mathrm{d}x}=1-4x^{-2}$ .

Key termsexpand first

Common mistake

Differentiating each bracket and multiplying: $\frac{\mathrm{d}}{\mathrm{d}x}\left[(x^{2}-3)(2x+1)\right]\neq(2x)(2)$ .

Section 5

Using the derivative

$f'(a)$ gives the gradient of the curve at $x=a$ and the rate of change of $f$ there.

To find the gradient at a point: substitute its $x$ -coordinate into $f'(x)$ .
To find where the gradient has a given value $m$ : solve $f'(x)=m$ , then substitute each $x$ into $f(x)$ (not $f'(x)$ ) for the $y$ -coordinate.

In economics, if $T(x)$ is total cost, $T'(x)$ is the marginal cost, the approximate cost of making one more item.

Key termsmarginal cost

Common mistake

Finding the $y$ -coordinate by substituting into $f'(x)$ . The $y$ -coordinate always comes from the original function.

Must know

$ax^{n}\to anx^{n-1}$ for any integer $n$ ; constants differentiate to 0.
Differentiate term by term.
Rewrite $\frac{a}{x^{n}}$ as $ax^{-n}$ ; the power goes more negative.
Expand products and split quotients before differentiating.
Gradient at a point: $f'(a)$ . Points with a given gradient: solve $f'(x)=m$ , then use $f(x)$ for $y$ .

That's the notes covered.

Carry on to the next subtopic.