Newton's Quotient: Derivative Definition

The Slope of the Tangent

Early mathematicians such as Isaac Newton were searching for a way to find the slope of the tangent at any point on a curve, since they realized that the tangents to the curve define its shape. Once they had mastered the concept of limits, they were able to find an expression to represent the slope of any tangent to any curve at any given point.

Here's what they did:

Since slope is the ratio of rise to run, they first found an expression for the slope of a secant to a curve y = f (x). Then they realized that the only difference between a secant and a tangent is that a secant intersects the curve at 2 points, and a tangent touches the curve at a single point. From the diagram, we see that if we collapse the horizontal distance h between the 2 points of contact, so that it is extremely close to zero, the line will become a tangent to the curve. This is accomplised by limiting h to zero. Some books use instead of h to represent the change in the x-value.

 The slope of the tangent to y = f(x) isThis is the definition of the derivative, also known as Newton's Quotient.

If asked to find the derivative of a function by first principles or by definition, we use this quotient. If we want the general derivative, we find an expression in x. If we need a specific slope or derivative, we evaluate the derivative expression at the specified value of x.

Example 1: Find an expression for the slope of the tangent to f (x) = 2x² + x – 3 at any point.

Solution: We want to find

When we remove brackets and collect like terms, we get:

When we cancel the h's in top and bottom and limit it to zero (0), we get f '(x) = 4x + 1

When we're given a specific point on the curve and asked to find the slope of the tangent at that point, we substitute the x-value of the point into the expression. Say we want the slope of the tangent to this curve at the point (2, 7), we simply set x = 2 to get f '(2) = 4(2) + 1 = 9

Note: x is a dummy variable. This means that it can be replaced with any variable, so to specify the derivative at the point (a, f (a)) we replace the x in the formula with a.

Example 2: Find the equation of the tangent to the curve in example 1 at the point where x = 3.

Solution: To find the equation of a line, we need a point and a slope. Since we know that x = 3, we can find f (3) to get the y-value at the point. It is 18 so the point is (3, 18). Now we know that the derivative is 4x + 1, so the slope of this tangent is 4(3) + 1 = 13.

The equation of the tangent to f (x) = 2x² + x – 3 at the point (3, 18) is:

or y = 13x – 21

Example 3: Find the equation of the tangent to at the point where x = 8.

Solution: When x = 8, y = 5, so the point is (8, 5) -- all we need now is the slope. Since the slope of the tangent is the derivative, found using Newton's Quotient, we find the numerical value of:

at x = 8

To do this, we will rationalize the numerator of this fraction.

Now we cancel the h top and bottom and set it = 0. We get:

The slope of the tangent is 3/10 so the equation is

Now we do one with a dreaded fraction.

Example 4: Find an algebraic expression for the derivative of

Solution:

Alternative Definition of the Derivative at x = a

Should we want the derivative at x = a, we can define the derivative as the slope between any point on the curve (x, f(x)) and the point in question (a, f(a)). In such a case, the horizontal distance between the points is no longer h. Now it is x – a, so instead of limiting h to zero, we make x approach a to collapse the interval. Now we can define the derivative as

Example 5: Given f (x) = x 1/ 3, find f '(a) using the alternative definition for the derivative.

Solution: Substituting f (x) = x 1/ 3, we get:

Now we consider x to be the cube of x 1/ 3 and a to be the cube of a 1/ 3 and we factor the difference of cubes in the denominator like this:

Now we divide by ( x 1/ 3a 1/ 3 ) and we make x approach a. So we end up with

Practice

1) Use Newton's Quotient to find an algebraic expression for the derivative of:

 a) f (x) = 2 + 8x – 5x² b) f (x) = x³ + x c) f (x) = 1 / (x + 5)

2) Write the equation of the tangent to functions in question 1 at the point where x = 3.

3) Use the alternative definition of the derivative to find f ' (a) if f (x) = x 4 + 2.

4) Find the point(s) on f (x) = x² – 6x + 3 where the tangent is horizontal. (hint: slope = 0).

Solutions

1) a)

b)

c)

2) a) The point is (3, –19) and the slope is –22, so the equation of the tangent is y = –22x + 47.

b) The point is (3, 30) and the slope is 28, so the equation of the tangent is y = 28x – 54.

c) The point is (3, 1/8) and the slope is –1/64, so the equation of the tangent is .

3) Use the alternative definition of the derivative to find f ' (a) if f (x) = x 4 + 2.

4) Find the point(s) on f (x) = x² – 6x + 3 where the tangent is horizontal. (hint: slope = 0).

Since the derivative must = 0 for the tangent to be horizontal, and f ' (x) = 2x – 6, we solve for x from 2x – 6 = 0. We find that x = 3, so y = – 6. The point is (3, – 6). This of course is the vertex of the parabola -- the only point where the tangent is horizontal.

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