MCV4U Power, Constant, Sum and Difference Rules: Notes and Practice

In the last unit you found derivatives from first principles: write $\frac{f ( x + h ) - f ( x )}{h}$ , expand, simplify, cancel the $h$ , then let $h \to 0$ . It works, but finding the derivative of something like $f (x) = 4 x^{5} - 3 x^{2} + 7 x$ that way means expanding $(x + h)^{5}$ and grinding through a page of algebra. Mathematicians did that grinding once, noticed that the answers follow a very simple pattern, and turned the pattern into rules. This lesson is about those rules, and about why you can trust them. Here is first principles for $f (x) = x^{3}$ . Expanding $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3},$ the $x^{3}$ terms cancel, every remaining term has a factor of $h$ , and dividing by $h$ leaves something you can safely let $h$ shrink in:

\frac{( x + h ) ^{3} - x ^{3}}{h} = \frac{3 x ^{2} h + 3 x h ^{2} + h ^{3}}{h} = 3 x^{2} + 3 x h + h^{2} ⟶ 3 x^{2} as h \to 0

Only the term with a single factor of h survives the limit; its coefficient, 3x², is the old exponent times x to one less power.

Do the same for $x$ , $x^{2}$ and $x^{4}$ and you get $1$ , $2 x$ and $4 x^{3}$ . Every time, the old exponent comes down in front as a multiplier and the new exponent is one less. That is the POWER RULE. The surviving term always comes from the second term of the binomial expansion of $(x + h)^{n}$ , which is $n x^{n - 1} h$ , so the pattern holds for every natural number $n$ .

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

The power rule. It holds for every real exponent n; in this course you use it for natural and rational exponents, including negative ones.

Fig 1A table of functions and their derivatives: x, x squared, x cubed and x to the fourth, the general x to the n highlighted, and the rational-exponent cases x to the one-half and x to the negative one in a dashed box below.

Constant, constant multiple, sum and difference rules

Rule	In symbols	Example
Constant	$\frac{d}{d x} (c) = 0$	$\frac{d}{d x} (7) = 0$
Power	$\frac{d}{d x} (x^{n}) = n x^{n - 1}$	$\frac{d}{d x} (x^{6}) = 6 x^{5}$
Constant multiple	$\frac{d}{d x} [k f (x)] = k f^{'} (x)$	$\frac{d}{d x} (4 x^{3}) = 4 (3 x^{2}) = 12 x^{2}$
Sum	$(f + g)^{'} (x) = f^{'} (x) + g^{'} (x)$	$\frac{d}{d x} (x^{2} + x^{5}) = 2 x + 5 x^{4}$
Difference	$(f - g)^{'} (x) = f^{'} (x) - g^{'} (x)$	$\frac{d}{d x} (x^{3} - 8 x) = 3 x^{2} - 8$

Together these differentiate any polynomial term by term.

Constant rule: the graph of $y = c$ is a horizontal line, and a horizontal line has slope $0$ everywhere.
Constant multiple rule: $y = k f (x)$ is a vertical stretch of $y = f (x)$ by a factor of $k$ . Every rise is multiplied by $k$ while every run stays the same, so every tangent slope is multiplied by $k$ too.
Sum and difference rules: if water flows into one barrel at $f^{'} (t)$ litres per minute and into a second barrel at $g^{'} (t)$ litres per minute, the total amount $f (t) + g (t)$ grows at $f^{'} (t) + g^{'} (t)$ litres per minute. Rates of change add, just as the amounts do.

\frac{d}{d x} [k f (x)] = h \to 0 lim \frac{k f ( x + h ) - k f ( x )}{h} = k h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = k f^{'} (x)

Reading the proof of the constant multiple rule: factor the k out of the difference quotient, and it passes straight through the limit.

VideoJensenMath works through the power rule together with the constant multiple and sum rules to differentiate polynomials term by term, as this lesson does.Watch on YouTube

Rational and negative exponents

Rewrite as a power of x before you differentiate

The power rule only sees terms of the form

k x^{n}

. Turn radicals into fractional exponents and fractions into negative exponents first:

x = x^{\frac{1}{2}}

3 x^{2} = x^{\frac{2}{3}}

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

, and

\frac{4}{x ^{2}} = 4 x^{- 2}

. Then bring the exponent down and subtract one, exactly as before.

Does the rule really work for an exponent like $\frac{1}{2}$ ? Check it numerically. The power rule says $\frac{d}{d x} (x^{\frac{1}{2}}) = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 x}$ , so at $x = 4$ the tangent slope should be $\frac{1}{2 ( 2 )} = 0.25$ . The table shows secant slopes $\frac{4 + h - 2}{h}$ closing in on exactly that value.

h	Secant slope of the square root function from x = 4 to x = 4 + h
$0.1$	$0.24846$
$0.01$	$0.24984$
$0.001$	$0.24998$
Power rule	$0.25$

Secant slopes rounded to 5 decimal places. They approach the power rule's prediction as h shrinks.

Once you have $f^{'} (x)$ you can answer two kinds of question quickly. Forward: what is the instantaneous rate of change (the slope of the tangent) at $x = a$ ? Substitute $a$ into $f^{'} (x)$ . Backward: where does a given rate of change happen? Set $f^{'} (x)$ equal to that rate and solve for $x$ , then substitute into $f (x)$ (not $f^{'} (x)$ ) to get the point. A tangent line at $x = a$ then has equation $y - f (a) = f^{'} (a) (x - a)$ .

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Pick a polynomial, drag a point along it, and watch the tangent slope plot the derivative point by point.

The power rule needs the variable in the base

The power rule applies to

x^{n}

, where the base is the variable and the exponent is a fixed number. It does NOT apply to

2^{x}

, where the variable is in the exponent. Exponential functions have their own rule, which you meet at the end of this unit.

Power, Constant, Sum and Difference Rules

Key things to remember

Constant, constant multiple, sum and difference rules

Rational and negative exponents

Worked examples

Where marks get lost

Flashcards 8

Practice