The Calculus Primer (2011)
Part II. The Derivative of a Function
Chapter 7. DIFFERENTIATION: FINDING THE DERIVATIVE
2—7. The Process of Differentiation. The operation of finding the derivative of a function is called differentiation. The derivative, as we have already seen, is written . This symbol is read: “the derivative of y with respect to x,” or, sometimes, the “x-derivative of y” It may be written in a variety of ways, but they all denote the same thing: thus
orf(x) = f′(x).
We shall now illustrate the operation of differentiation, and then formulate a General Rule of procedure.
EXAMPLE 1.Differentiate: y = 3x − 2.
Solution.
y = 3x − 2.(1)
Let x take on an increment Δx, and let the corresponding increment in y be Δy.
Theny + Δy = 3(x + Δx) − 2,
ory + Δy = 3x + 3Δx − 2.(2)
Subtract equation (1) from equation (2):
y + Δy = 3x + 3Δx − 2
y= 3x − 2
Δy = 3Δx(3)
Now divide equation (3) by Δx:
Finally, let Δx → 0:
= 3,
or = 3.
EXAMPLE 2.Differentiate: y = x2 + 5.
Solution.
y = x2 + 5.(1)
y + Δy = (x + Δx)2 + 5,
ory + Δy = x2 + 2x·Δx+ (Δx)2 + 5.(2)
Subtracting (1) from (2):
y + Δy = x2 + 2x·Δx + (Δx)2 + 5
y= x2+ 5
Δy = 2x·Δx + (Δx)2(3)
Dividing (3) by Δx:
= 2x + Δx.
Let Δx → 0:
= 2x,
or = 2x.
EXAMPLE 3.Differentiate: y = x2 + 3x + 8.
Solution.
y + Δy = (x + Δx)2 + 3(x + Δx) + 8
y + Δy = x2 + 2x·Δx + (Δx)2 + 3x + 3Δx + 8
y= x2+ 3x+ 8
Δy = 2x·Δx + (Δx)2 + 3Δx
Dividing through by Δx:
= 2x + Δx + 3.
Let Δx → 0:
= 2x + 3.
or = 2x + 3.
EXAMPLE 4.Differentiate: y = x3.
Solution.y + Δy = (x + Δx)3.
y + Δy = x3 + 3x2·Δx + 3x·(Δx)2 + (Δx)3.
Subtracting:Δy = 3x2·Δx + 3x (Δx)2 + (Δx)3.
Dividing: = 3x2 + 3x(Δx) + (Δx)2.
Passing to the limit:
= 3x2,
for, as Δx → 0, the terms 3x(Δx) and (Δx)2 vanish.
EXAMPLE 5.Differentiate: y = .
2—8. The General Rule for Differentiation. We may now formulate the “four-step” rule for differentiating a function, as follows:
Step 1. Substitute (x + Δx) for x in the given equation, thus giving y a new corresponding value, (y + Δy).
Step 2. Subtract the given equation from the equation obtained in Step 1, thus obtaining an expression for Δy.
Step 3. Divide the equation obtained in Step 2 by Δx, thus obtaining a value for .
Step 4. Find the limit of as Δx approaches zero as a limit. This limit is the required derivative.
EXERCISE 2—1
Differentiate each of the following functions by the General Rule:
1. y = 2x − 3
2. y = 4 − x2
3. y = 2x2 + 3x
4. y = 10x2
5. y = x2 − x + 6
6. y = 3x2 − 2x + 1
7. y = 2x3 + 3
8. y = x3 − 2x
9. y = (x + 1)(x − 2)
10. y = (2x − l)(3x + 2)
11.y =
12.y =
13.y =
14.y =
15.y =
16.y =
2—9. Finding the Tangent to a Curve. In the light of the discussion in §2—6, we are now able to find the slope of the tangent to a given curve at a given point, or the slope of a curve at any desired point, provided, of course, that the equation of the curve is given.
EXAMPLE 1.Find the slope of the curve y = x2 + 6 at the point where x = 5.
Solution.y = x2 + 6.
By the General Rule,
= 2x + Δx,
and = 2x.
Hence, when x = 5, = 2(5) = 10. The slope of the tangent to y = x2 + 6 at the point where x = 5 is 10, as is the slope of the curve at that point. The inclination of the tangent is ϕ = arc tan 10, or approximately 84°17′.
EXAMPLE 2.Find the slope of the curve y = 3x2 − 4x + 8 at the point where x = 3.
Solution.By the General Rule,
= 6x − 4 + 3(Δx),
and = 6x − 4.
Hence, when x = 3, = 6(3) − 4 = 14, the required slope.
EXERCISE 2—2
By differentiating by the General Rule, find the slope of the tangent to each of the following curves at the point indicated:
1. y = x2 − 5, where x = 6.
2. y = 4x2 + 3, where x = 1.
3. y = x2 − 3x + 6, where x = 4.
4. y = 3x2 + 4x, where x = − 2.
5. y = (x + 3)2, where x = 0.
6. y = x3 − x2, where x = −3.
7. y = x3 + 8, where x = −2.
8. y = x4 − 1, where x = .