Year 12 Maths - Methods (Unit 3 & Unit 4) Differentiation

Rules for differentiation

Q: What is the power rule for differentiation?

The power rule says d/dx of x to the n equals n times x to the n minus 1: bring the index down as a multiplier and reduce it by one. For example d/dx of x to the 5 equals 5 x to the 4.

Q: How do you differentiate a polynomial?

Differentiate a polynomial one term at a time. Apply the power rule to each power of x, carry constant multiples through, and the constant term differentiates to zero. So d/dx of 4 x cubed plus 2 x squared minus 5x plus 8 equals 12 x squared plus 4x minus 5.

Q: What is the derivative of a constant?

The derivative of a constant is zero, because a constant function is a horizontal line with gradient zero. So d/dx of 8 equals 0, and the constant term of a polynomial disappears when you differentiate.

Q: What is the derivative of x, and of a term like 5x?

Since x equals x to the 1, the power rule gives d/dx of x equals 1. A constant multiple is carried through, so d/dx of 5x equals 5; in general a linear term ax differentiates to a.

Q: How do you find the gradient of a curve at a point?

Differentiate the function to get f prime of x, then substitute the x value of the point. The number f prime of a is the gradient of the curve, and of the tangent, at x equals a. For f of x equals x squared minus 4x plus 3, f prime of x equals 2x minus 4, so f prime of 3 equals 2.

Q: How do you find the equation of a tangent to a polynomial?

First find the gradient m equal to f prime of a at the point, then use y minus y1 equals m times x minus x1 with the point on the curve. For y equals x squared minus 4x plus 3 at the point (3,0), the gradient is 2 and the tangent is y equals 2x minus 6.

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Theory

Differentiating a polynomial uses three rules: the power rule \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\), the constant-multiple rule, and the sum/difference rule — so a polynomial is differentiated one term at a time. The derivative of a constant is \(0\), and evaluating \(f'(a)\) gives the gradient of the curve at that point.

The derivative \(f'(x)\) (also written \(\dfrac{dy}{dx}\)) gives the gradient of the curve \(y=f(x)\) at each value of \(x\). For a positive-integer power of \(x\), the power rule applies: bring the index down as a multiplier and reduce it by one, \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\). For example \(\dfrac{d}{dx}x^{5}=5x^{4}\), and since \(x=x^{1}\), \(\dfrac{d}{dx}x=1\).

A constant multiple is carried straight through, \(\dfrac{d}{dx}\big(a\,f(x)\big)=a\,f'(x)\), and sums and differences are differentiated term by term, \(\dfrac{d}{dx}\big(f(x)\pm g(x)\big)=f'(x)\pm g'(x)\). Together these let you differentiate any polynomial one term at a time: \(\dfrac{d}{dx}(a x^{n})=a n\,x^{n-1}\).

The derivative of a constant is \(0\), because a constant function is a horizontal line with zero gradient — so the constant term of a polynomial disappears when you differentiate. Substituting \(x=a\) into \(f'(x)\) gives the number \(f'(a)\), the gradient of the tangent to \(y=f(x)\) at \(x=a\); the tangent line then follows from \(y-y_{1}=m(x-x_{1})\).

Key idea. \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\). Differentiate a polynomial term by term, carrying constant multiples through and sending each constant term to \(0\); then \(f'(a)\) is the gradient at \(x=a\).

\(y=x^{2}\): the derivative \(\dfrac{dy}{dx}=2x\) gives gradient \(2\) at \((1,1)\), so the tangent is \(y=2x-1\)

\(y=x^{2}-2x\): here \(\dfrac{dy}{dx}=2x-2\), so \(f'(1)=0\) — a horizontal tangent at the turning point \((1,-1)\)

The power rule for a positive-integer index — multiply by the index, then reduce it by one:

\[\frac{d}{dx}x^{n}=n\,x^{n-1}\]

\frac{d}{d x} x^{n} = n x^{n - 1}

The constant-multiple rule and the sum/difference rule — differentiate a polynomial term by term:

\[\frac{d}{dx}\big(a\,x^{n}\big)=a n\,x^{n-1} \qquad \frac{d}{dx}\big(f(x)\pm g(x)\big)=f'(x)\pm g'(x)\]

\frac{d}{d x} (a x^{n}) = a n x^{n - 1}

The derivative of a constant is zero, and evaluating the derivative at \(x=a\) gives the gradient there:

\[\frac{d}{dx}(c)=0 \qquad m=f'(a)=\left.\frac{dy}{dx}\right|_{x=a}\]

\frac{d}{d x} (c) = 0

Gradient of a tangent. The derivative at \(x=a\) is the gradient there; e.g. for \(y=x^{2}-4x+3\), \(\dfrac{dy}{dx}=2x-4\), so at \(x=3\) the gradient is \(2\) and the tangent is \(y=2x-6\).

How to differentiate a polynomial

Write each term as a power of \(x\). Put every term in the form \(a\,x^{n}\); read a bare \(x\) as \(x^{1}\) and a lone number as the constant term.
Apply the power rule term by term. For each \(a\,x^{n}\), multiply by the index and reduce it by one to get \(a n\,x^{n-1}\).
Handle the constant term. A constant differentiates to \(0\), so it simply drops out; a linear term \(ax\) becomes \(a\).
Combine. Add the term derivatives to write \(f'(x)\), the derivative of the whole polynomial.
Gradient or tangent. To find a gradient, substitute the \(x\)-value into \(f'(x)\); for a tangent line use \(y-y_{1}=m(x-x_{1})\) with \(m=f'(a)\) and the point.

Quick check. Differentiating a degree-\(n\) polynomial gives a degree-\((n-1)\) polynomial, so a cubic \(\to\) quadratic, a quadratic \(\to\) linear, and a constant \(\to 0\).

Example 1 — power rule and a constant

Differentiate \(y=x^{6}\) and \(y=7\).

Solution

\(y=x^{6}\)

Power rule — multiply by the index \(6\), then reduce it by one:

\(\dfrac{dy}{dx}\)	\(=\)	\(6\,x^{6-1}\)
	\(=\)	\(6x^{5}\)

\(\therefore\) \(\dfrac{dy}{dx}=6x^{5}\)

\(y=7\)

A constant is a horizontal line — its gradient is \(0\):

\(\dfrac{dy}{dx}\)

\(=\)

\(0\)

\(\therefore\) \(\dfrac{dy}{dx}=0\)

6 x^{5}

Example 2 — polynomial term by term

Differentiate \(y=4x^{3}+2x^{2}-5x+8\).

Solution

Differentiate each term — carry the constant multiples through:

\(\dfrac{d}{dx}(4x^{3})\)	\(=\)	\(12x^{2}\)
\(\dfrac{d}{dx}(2x^{2})\)	\(=\)	\(4x\)
\(\dfrac{d}{dx}(-5x)\)	\(=\)	\(-5\)
\(\dfrac{d}{dx}(8)\)	\(=\)	\(0\)

Combine the term derivatives:

\(\dfrac{dy}{dx}\)

\(=\)

\(12x^{2}+4x-5\)

\(\therefore\) \(\dfrac{dy}{dx}=12x^{2}+4x-5\)

12 x^{2} + 4 x - 5

Example 3 — evaluating \(f'(a)\)

For \(f(x)=x^{3}-4x+1\), find \(f'(2)\).

Solution

Differentiate term by term — the constant \(1\) drops out:

\(f'(x)\)

\(=\)

\(3x^{2}-4\)

Substitute \(x=2\):

\(f'(2)\)	\(=\)	\(3(2)^{2}-4\)
	\(=\)	\(12-4=8\)

\(\therefore\) \(f'(2)=8\) — the gradient at \(x=2\)

f^{'} (2) = 8

Example 4 — gradient and tangent at a point

For \(y=x^{2}-4x+3\), find the gradient at \(x=3\) and the equation of the tangent there.

Solution

Gradient — differentiate, then substitute \(x=3\):

\(\dfrac{dy}{dx}\)	\(=\)	\(2x-4\)
\(m=\left.\dfrac{dy}{dx}\right\|_{x=3}\)	\(=\)	\(2(3)-4=2\)

Point on the curve — substitute \(x=3\) into \(y\):

\(y\)

\(=\)

\((3)^{2}-4(3)+3=0\)

Tangent line — use \(y-y_{1}=m(x-x_{1})\) at \((3,0)\):

\(y-0\)	\(=\)	\(2(x-3)\)
\(y\)	\(=\)	\(2x-6\)

\(\therefore\) gradient \(=2\); tangent: \(y=2x-6\)

y = 2 x - 6

Common pitfalls

Reduce the index by one. The power rule lowers the power: \(\dfrac{d}{dx}x^{4}=4x^{3}\), not \(4x^{4}\). Multiply by the old index, then subtract one from it.

The derivative of a constant is \(0\), and a linear term loses its \(x\). \(\dfrac{d}{dx}(8)=0\) (not \(8\)), and \(\dfrac{d}{dx}(5x)=5\) — the \(x\) disappears because \(x^{1}\to x^{0}=1\).

Differentiate every term, and keep signs. Work term by term across the whole polynomial; a minus sign in front of a term stays with its derivative, so \(\dfrac{d}{dx}(-5x)=-5\).

Frequently asked questions

What is the power rule for differentiation?

The power rule: \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\). Bring the index down as a multiplier and reduce it by one, e.g. \(\dfrac{d}{dx}x^{5}=5x^{4}\).

How do you differentiate a polynomial?

Differentiate one term at a time — apply the power rule to each term, carry constant multiples through, and the constant term becomes \(0\). So \(\dfrac{d}{dx}(4x^{3}+2x^{2}-5x+8)=12x^{2}+4x-5\).

What is the derivative of a constant?

Zero. A constant is a horizontal line with gradient \(0\), so \(\dfrac{d}{dx}(c)=0\) and the constant term disappears when you differentiate.

What is the derivative of x, and of a term like 5x?

Since \(x=x^{1}\), \(\dfrac{d}{dx}x=1\); a constant multiple is carried through, so \(\dfrac{d}{dx}(5x)=5\). In general \(\dfrac{d}{dx}(ax)=a\).

How do you find the gradient of a curve at a point?

Differentiate, then substitute the point: \(f'(a)\) is the gradient at \(x=a\). For \(f(x)=x^{2}-4x+3\), \(f'(x)=2x-4\), so \(f'(3)=2\).

How do you find the equation of a tangent to a polynomial?

Find the gradient \(m=f'(a)\), then use \(y-y_{1}=m(x-x_{1})\). For \(y=x^{2}-4x+3\) at \((3,0)\), \(m=2\) and the tangent is \(y=2x-6\).

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Rules for differentiation

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How to differentiate a polynomial

Common pitfalls

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