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Introduction to differentiation
Introduction to differentiation

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2.2 Constant multiple rule

In this section and the next, you’ll see two ways in which you can use formulas that you know for the derivatives of functions to find formulas for the derivatives of other, related functions.

First, suppose that you know the formula for the derivative of a particular function, and you want to know the formula for the derivative of a constant multiple of the function. For example, you already know the formula for the derivative of the function f of x equals x squared , but suppose that you want to know the formula for the derivative of the function g of x equals three times x squared . Let’s think about how the formula for the derivative of the second function can be worked out from the formula for the derivative of the first function.

When you multiply a function by a constant, the effect on its graph is that, for each x -value, the corresponding y -value is multiplied by the constant. So the graph is stretched or squashed vertically, and, if the constant is negative, then it’s also reflected in the x -axis. These effects are called vertical scalings. For example, Figure 24 shows the graphs of y equals x squared , y equals three times x squared , y equals one divided by two times x squared and y equals negative x squared , and the point with x -coordinate 1 on each of these graphs.

You can see the following effects.

  • Multiplying the function f of x equals x squared by the constant 3 scales its graph vertically by a factor of 3 (which stretches it).

  • Multiplying the function f of x equals x squared by the constant one divided by two scales its graph vertically by a factor of one divided by two (which squashes it).

  • Multiplying the function f of x equals x squared by the constant negative one scales its graph vertically by a factor of negative one (which reflects it in the x -axis).

Described image
Figure 24 The graphs of (a)  y equals x squared (b)  y equals three times x squared (c)  y equals one divided by two times x squared (d)  y equals negative x squared
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The graphs of (ay) y = x squared, (b) y = 3 times x squared, (c) y = one half of x squared, (d) y = minus x squared. Each graph is plotted on a grid of unit squares. Both axes are shown, but there are no scales on them. Graph (ay) shows the u shaped parabola y = x squared. It passes through the origin and is symmetric about the positive y-axis. The point (1, 1) is marked with a dot on the curve. Graph (b) shows the u shaped parabola y = 3 times x squared. It passes through the origin and is symmetric about the positive y-axis. It rises more steeply than the parabola y = x squared. The point (1, 3) is marked with a dot on the curve. Graph (c) shows the curve y = one half of x squared. This is also a u shaped parabola which passes through the origin and is symmetric about the positive y-axis. It rises less steeply than the parabola y = x squared. The point (1, 0.5) is marked with a dot on the curve. Graph (d) shows the curve y = minus x squared. This is an n shaped parabola which passes through the origin and is symmetric about the negative y-axis. It resembles the parabola y = x squared after reflection in the x-axis. The point (1, minus 1) is marked with a dot on the curve.

Figure 24 The graphs of (a)  y equals x squared (b)  y equals three times x squared (c)  y equals one divided by two times x squared (d)  y equals negative x squared

The stretching or squashing, and possible reflection, of the graph causes the gradient at each x -value to change. For example, you can see that, at the point with x -coordinate 1, the graph of y equals three times x squared is steeper than the graph of y equals x squared .

To see exactly how the gradients change, first consider what happens to the gradient of a straight line when you scale it vertically by a particular factor, say a . The scaled line will go up by a times as many units for every one unit that it goes along, compared to the unscaled line. In other words, its gradient is multiplied by the factor a . For example, Figure 25(a) illustrates what happens when you take a straight line with gradient 1 and scale it vertically by a factor of 3.

The same thing happens for any graph: if you scale it vertically by a particular factor, then its gradient at any particular x -value is multiplied by this factor. For example, Figure 25(b) illustrates that if you take a curve that has gradient 1 at a particular x -value, and scale it vertically by a factor of 3, then the new curve has gradient 3 at that x -value.

Described image
Figure 25 (a) A straight line with gradient 1, and the result of scaling it vertically by a factor of 3 (b) a curve that has gradient 1 at a particular x -value, and the result of scaling it vertically by a factor of 3
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A straight line with gradient 1, and the result of scaling it vertically by a factor of 3. (b) A curve that has gradient 1 at a particular x value, and the result of scaling it vertically by a factor of 3. There are two graphs in each part. Each graph is drawn on a grid of squares. The x and y axes are shown but there are no scales on them. In the first graph in (ay) a straight line cuts the negative x-axis one square to the left of the origin and the positive y-axis one square above the origin. In the second graph in (ay) a straight line cuts the negative x-axis one square to the left of the origin and the positive y-axis three squares above the origin. In the first graph in (b) a curve is drawn which is symmetric about the y-axis. It cuts the y-axis one and a half squares above the origin and rises in a smooth curve either side of this point. A dot is marked on the curve at the point which is one square to the right of the origin and two squares up. A tangent is drawn to touch the curve at this point. When extended downwards it cuts the y-axis one square up from the origin and the x-axis one square to the left of the origin. In the second graph in (b) a curve is drawn which is symmetric about the y-axis. It cuts the y-axis four and a half squares above the origin and rises in a smooth curve either side of this point. A dot is marked on the curve at the point which is one square to the right of the origin and six squares up. A tangent is drawn to touch the curve at this point. When extended downwards it cuts the y-axis three squares up from the origin and the x-axis one square to the left of the origin.

Figure 25 (a) A straight line with gradient 1, and the result of scaling it vertically by a factor of 3 (b) a curve that has gradient 1 ...

So, if you multiply a function by a constant, then its derivative is multiplied by the same constant. This fact can be stated as in the box below.

Constant multiple rule (Lagrange notation)

If the function k is given by k of x equals a times f of x , where f is a function and a is a constant, then

k super prime of x equals a times f super prime of x comma

for all values of x at which f is differentiable.

For example, since the derivative of f of x equals x squared is f super prime of x equals two times x , it follows by the constant multiple rule that

  • the derivative of g of x equals three times x squared is equation sequence part 1 g super prime of x equals part 2 three multiplication two times x equals part 3 six times x

  • the derivative of h of x equals one divided by two times x squared is equation sequence part 1 h super prime of x equals part 2 one divided by two multiplication two times x equals part 3 x

  • the derivative of p of x equals negative x squared is p super prime of x equals negative two times x .

The third of these follows because taking a negative is the same as multiplying by  negative one . It’s useful to remember in general that if f and k are functions such that k of x equals negative f of x , then, by the constant multiple rule,

k super prime of x equals negative f super prime of x comma

for all values of x at which f is differentiable. Like everything involving derivatives, the constant multiple rule can also be stated in Leibniz notation, as follows.

Constant multiple rule (Leibniz notation)

If y equals a times u , where u is a function of x and a is a constant, then

d y divided by d x equals a times d u divided by d x comma

for all values of x at which u is differentiable.

(The phrase ‘ u is differentiable’ in the box above is a condensed way of saying that if we write u equals f of x then f is differentiable at  x .)

The constant multiple rule can be proved formally by using the idea of differentiation from first principles, and you’ll see this done at the end of this section. First, however, you should concentrate on learning to use it. Here’s an example.

Example 3 Using the constant multiple rule

Differentiate the following functions.

  • a. f of x equals eight times x super four

  • b. f of x equals negative Square root of x

  • c. y equals three divided by x

Solution

  • a.The derivative is 8 times the derivative of x super four .

    equation sequence part 1 f super prime of x equals part 2 eight multiplication four times x cubed equals part 3 32 times x cubed

  • b.The derivative is the negative of the derivative of Square root of x .

    f of x equals negative x super one solidus two , so

    equation sequence part 1 f super prime of x equals part 2 negative one divided by two times x super negative one solidus two equals part 3 negative one divided by two multiplication one divided by x super one solidus two equals part 4 negative one divided by two times x super one solidus two equals part 5 negative one divided by two times Square root of x

  • c.The derivative is 3 times the derivative of one solidus x .

    y equals three times x super negative one , so

    equation sequence part 1 d y divided by d x equals part 2 three multiplication left parenthesis negative one right parenthesis times x super negative two equals part 3 negative three divided by x squared

Here are some examples for you to try. Notice that in some of them the letters used are not the standard ones, x , y and f .

Activity 8 Using the constant multiple rule

Differentiate the following functions.

  • a. f of x equals five times x cubed

  • b. f of x equals negative x super seven

  • c. f of x equals two times Square root of x

  • d. f of x equals six times x

  • e. f of x equals x divided by four

  • f. f of t equals two divided by t

  • g. g of u equals negative seven times u

  • h. y equals Square root of x divided by three

  • i. y equals eight divided by x squared

  • j. y equals negative five divided by x

  • k. y equals four divided by Square root of x

  • l. y equals four times x super three solidus two

  • m. p equals one divided by three times Square root of u

  • n. q equals negative 12 times r super one solidus three

Answer

  • a. f of x equals five times x cubed , so

    equation sequence part 1 f super prime of x equals part 2 five multiplication three times x squared equals part 3 15 times x squared full stop

  • b. f of x equals negative x super seven , so

    f super prime of x equals negative seven times x super six full stop

  • c. equation sequence part 1 f of x equals part 2 two times Square root of x equals part 3 two times x super one solidus two , so

    multiline equation row 1 f super prime of x equals two multiplication one divided by two times x super negative one solidus two row 2 Blank equals one divided by x super one solidus two row 3 Blank equals one divided by Square root of x full stop

  • d. f of x equals six times x , so

    equation sequence part 1 f super prime of x equals part 2 six multiplication one equals part 3 six full stop

  • e. equation sequence part 1 f of x equals part 2 x divided by four equals part 3 one divided by four times x , so

    equation sequence part 1 f super prime of x equals part 2 one divided by four multiplication one equals part 3 one divided by four full stop

  • f. equation sequence part 1 f of t equals part 2 two divided by t equals part 3 two times t super negative one , so

    multiline equation row 1 f super prime of t equals two multiplication left parenthesis negative one right parenthesis times t super negative two row 2 Blank equals negative two divided by t squared full stop

  • g. g of u equals negative seven times u , so

    equation sequence part 1 g super prime of u equals part 2 negative seven multiplication one equals part 3 negative seven full stop

  • h. equation sequence part 1 y equals part 2 Square root of x divided by three equals part 3 one divided by three times x super one solidus two , so

    multiline equation row 1 d y divided by d x equals one divided by three multiplication one divided by two times x super negative one solidus two row 2 Blank equals one divided by six multiplication one divided by x super one solidus two row 3 Blank equals one divided by six times Square root of x full stop

  • i. equation sequence part 1 y equals part 2 eight divided by x squared equals part 3 eight times x super negative two , so

    multiline equation row 1 d y divided by d x equals eight multiplication left parenthesis negative two right parenthesis times x super negative three row 2 Blank equals negative 16 divided by x cubed full stop

  • j. equation sequence part 1 y equals part 2 negative five divided by x equals part 3 negative five times x super negative one , so

    multiline equation row 1 d y divided by d x equals negative five multiplication left parenthesis negative one right parenthesis times x super negative two row 2 Blank equals five divided by x squared full stop

  • k. equation sequence part 1 y equals part 2 four divided by Square root of x equals part 3 four divided by x super one solidus two equals part 4 four times x super negative one solidus two , so

    multiline equation row 1 d y divided by d x equals four multiplication left parenthesis negative one divided by two right parenthesis times x super negative three solidus two row 2 Blank equals negative two divided by x super three solidus two full stop

  • l. y equals four times x super three solidus two , so

    multiline equation row 1 d y divided by d x equals four multiplication three divided by two times x super one solidus two row 2 Blank equals six times Square root of x full stop

  • m. equation sequence part 1 p equals part 2 one divided by three times Square root of u equals part 3 one divided by three times u super one solidus two equals part 4 one divided by three times u super negative one solidus two , so

    multiline equation row 1 d p divided by d u equals one divided by three multiplication left parenthesis negative one divided by two right parenthesis times u super negative three solidus two row 2 Blank equals negative one divided by six multiplication one divided by u super three solidus two row 3 Blank equals negative one divided by six times u super three solidus two full stop

  • n. q equals negative 12 times r super one solidus three , so

    multiline equation row 1 d q divided by d r equals negative 12 multiplication one divided by three times r super negative two solidus three row 2 Blank equals negative four multiplication one divided by r super two solidus three row 3 Blank equals negative four divided by r super two solidus three full stop

Activity 9 Using the constant multiple rule to find a gradient

Find the gradient of the graph of the function f of x equals three times x squared at the point with x -coordinate 2.

Answer

The derivative of the function f of x equals three times x squared is

equation sequence part 1 f super prime of x equals part 2 three multiplication two times x equals part 3 six times x full stop

So the gradient of this function f at the point with x -coordinate 2 is

six multiplication two equals 12 full stop

You saw in the previous section that the function

f of x equals one

has derivative

f super prime of x equals zero full stop

This fact, together with the constant multiple rule, tells you that if a is any constant, then the function

f of x equals a left parenthesis which is the same as f of x equals a multiplication one right parenthesis

has derivative

equation sequence part 1 f super prime of x equals part 2 a multiplication zero equals part 3 zero full stop

For example, the function f of x equals three has derivative f super prime of x equals zero .

This is as you would expect, because the graph of the function f of x equals a (which is illustrated in Figure 26, in the case where a is positive) is a straight line with gradient 0, which means that the gradient at every point on the graph is 0.

Described image
Figure 26 The graph of f of x equals a
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The graph of f of x = ay. The axes are shown but no scales are marked on them. A horizontal line cuts the y-axis above the origin at a point labelled ay.

Figure 26 The graph of f of x equals a

This fact about the derivative of a constant function can be stated as follows.

Derivative of a constant function

If a is a constant, then

d divided by d x times left parenthesis a right parenthesis equals zero full stop

To finish this section, here’s a formal proof of the constant multiple rule, using differentiation from first principles. It uses the Lagrange notation form of the constant multiple rule, which is repeated below.

Constant multiple rule (Lagrange notation)

If the function k is given by k of x equals a times f of x , where f is a function and a is a constant, then

k super prime of x equals a times f super prime of x comma

for all values of x at which f is differentiable.

A proof of the constant multiple rule

Suppose that f is a function and a is a constant. Consider the function k given by k of x equals a times f of x . Let x be any value at which f is differentiable. To find k super prime of x , you have to consider what happens to the difference quotient for  k at x , which is

k times left parenthesis x plus h right parenthesis minus k of x divided by h

(where h can be positive or negative but not zero), as h gets closer and closer to zero. Since k of x equals a times f of x , the difference quotient for k at x is equal to

a times f times left parenthesis x plus h right parenthesis minus a times f of x divided by h comma

which is equal to

a of f times left parenthesis x plus h right parenthesis minus f of x divided by h full stop

The expression in the large brackets is the difference quotient for f at x , so, as h gets closer and closer to zero, it gets closer and closer to f super prime of x . Hence the whole expression gets closer and closer to a times f super prime of x . In other words,

k super prime of x equals a times f super prime of x comma

which is the constant multiple rule.