Rules For Differentiation
Master the rules for differentiation for Queensland Year 11 Mathematical Methods (QCAA). Rather than working from first principles every time, these shortcuts let you differentiate power and polynomial functions quickly and reliably.
You will learn the rule for differentiating a power of x, use the linearity of the derivative to work term by term, and solve where the gradient of the tangent is zero to find horizontal tangents.
Every question with a fully worked solution.
- Rules For Differentiation - Video - Differentiation of x^n where n is a positive integer Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), once the derivative is known you differentiate quickly with rules: the power rule \(\dfrac{d}{dx}x^n=nx^{n-1}\), the constant-multiple rule and the sum/difference rule. This page applies them to polynomials term by term, rewrites products before differentiating, and evaluates \(f'(a)\) or solves \(f'(x)=0\).
The power rule says \(\dfrac{d}{dx}x^n=nx^{n-1}\) for a positive integer \(n\): multiply by the power, then reduce the power by one.
The constant-multiple rule says \(\dfrac{d}{dx}\big(cf(x)\big)=c\,f'(x)\), and the sum/difference rule (linearity) says \(\dfrac{d}{dx}\big(f\pm g\big)=f'\pm g'\). Together they let you differentiate a polynomial term by term. The derivative of a constant is \(0\).
Because \(f'(a)\) is the gradient of the tangent at \(x=a\), setting \(f'(x)=0\) locates the horizontal tangents of the curve.
Power rule (positive integer \(n\)):
Constant-multiple and sum/difference (linearity):
Differentiating a polynomial with the rules
- Rewrite first if needed: expand any product and write each term as a power of \(x\).
- Power rule on each term: multiply by the power, then subtract one from the power.
- Coefficients and constants: carry coefficients along; the derivative of a constant is \(0\).
- Evaluate or solve: substitute \(x=a\) for \(f'(a)\), or set \(f'(x)=0\) for horizontal tangents.
Power rule — bring the power down, then reduce it by one:
| \(\dfrac{dy}{dx}\) | \(=\) | \(3\times 4\,x^{4-1}\) |
| \(=\) | \(12x^3\) |
\(\dfrac{dy}{dx}=12x^3\).
Differentiate each term with the power rule:
| \(f'(x)\) | \(=\) | \(2(3x^2)-5(2x)+4(1)-0\) |
| \(=\) | \(6x^2-10x+4\) |
The constant \(-7\) differentiates to \(0\); the term \(4x\) gives \(4\).
\(f'(x)=6x^2-10x+4\).
Expand the product first (no product rule needed):
| \(y\) | \(=\) | \(x^2(3x-2)\) |
| \(=\) | \(3x^3-2x^2\) |
Now differentiate term by term:
| \(\dfrac{dy}{dx}\) | \(=\) | \(3(3x^2)-2(2x)\) |
| \(=\) | \(9x^2-4x\) |
\(\dfrac{dy}{dx}=9x^2-4x\).
Differentiate:
| \(f'(x)\) | \(=\) | \(3x^2-12x+9\) |
Horizontal tangent — solve \(f'(x)=0\):
| \(3x^2-12x+9\) | \(=\) | \(0\) |
| \(3(x^2-4x+3)\) | \(=\) | \(0\) |
| \(3(x-1)(x-3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(1,\ 3\) |
Gradient at \(x=2\) — substitute into \(f'\):
| \(f'(2)\) | \(=\) | \(3(2)^2-12(2)+9\) |
| \(=\) | \(12-24+9\) | |
| \(=\) | \(-3\) |
Horizontal tangents at \(x=1\) and \(x=3\); \(f'(2)=-3\).
Common pitfalls
Frequently asked questions
What is the power rule for differentiation?
For a positive integer \(n\), \(\dfrac{d}{dx}x^n=nx^{n-1}\): multiply by the power and reduce the power by one.
How do you differentiate a whole polynomial?
Differentiate each term separately with the power rule and add the results; this is the sum/difference and constant-multiple rules.
What is the derivative of a constant?
Zero. A constant term has no \(x\), so it does not change and \(\dfrac{d}{dx}(c)=0\).
Do you expand brackets before differentiating?
Yes, in Year 11. Expand a product such as \(x^2(3x-2)=3x^3-2x^2\) first, then differentiate term by term.
How do you find where a curve has a horizontal tangent?
Differentiate, set \(f'(x)=0\) and solve, because a horizontal tangent has gradient \(0\).