To differentiate a basic cubic function, we can go through the same process as we did to dif- ferentiate the basic linear and quadratic functions. Some of the expressions are a little more complicated this time, but otherwise everything works out in a similar way.
Cube x and multiply by constant
Let’s go through our trusty routine to find the derivative of a function f in which we cube the independent variable x and then multiply by the constant a, like this:
f (x ) =ax 3
As in previous situations, we’ll determine f ′ at a nonspecific point (x0,y0). In the xy-plane, we
can plot the graph of
y = ax 3
Thex-value of a movable point (x,y ) in the vicinity of (x0,y0) is, once again,
x =x0+ Δx
They-value of the movable point is y=ax 3=a(x
0+ Δx )3=a[x03+ 3x02Δx+ 3x0(Δx )2+ (Δx )3]
=ax03+ 3ax02Δx+ 3ax0(Δx )2+a(Δx )3
The slope of a line passing through (x0,y0) and (x,y ) is
Δy/Δx= ( y−y0) / (x−x0)
The function says that y0=ax03. In the above equation, let’s substitute
• The quantity [ax03+ 3ax02Δx+ 3ax0(Δx )2+a(Δx )3] for y
• The quantity ax03 for y0
That gives us Δy /Δx= [ax03+ 3ax02Δx+ 3ax0(Δx )2+a(Δx )3−ax03] / (x0+ Δx−x0) which simplifies to Δy /Δx= [3ax02Δx+ 3ax0(Δx )2+a(Δx )3] / Δx and finally to Δy /Δx= 3ax02+ 3ax0Δx+a(Δx )2
The derivative f ′(x0) is therefore
f ′(x0)= Δ →Limx 0 3ax02+ 3ax0Δx+a(Δx )2
AsΔx approaches 0, the second addend, 3ax0Δx, approaches 0 because it is a constant multiple
ofΔx. The third addend, a(Δx )2, also approaches 0. It’s a constant multiple of (Δx )2, which
must approach 0 as Δx approaches 0. The first addend, 3ax02, stays the same no matter what
Δx becomes. The limit of the entire expression is therefore 3ax02, showing that
f ′(x0)= 3ax02
for any real number x0. Therefore, any function of the form
f (x ) =ax 3
has the derivative
f ′(x ) = 3ax 2
Here’s a challenge!
Using the calculus techniques we’ve learned so far, find the derivative of the function
f (x ) = 2x 3−
5x
at the general point (x0,y0) in rectangular xy-coordinates, where y=f (x ).
Solution
We can plot this function as the graph of
y= 2x 3− 5x
We create the usual movable point (x,y ) near (x0,y0). The x-value of this point is x=x0+ Δx
They-value of the movable point is
y= 2x 3− 5x= 2(x
0+ Δx )3− 5(x0+ Δx )
= 2[x03+ 3x02Δx+ 3x0(Δx )2+ (Δx )3]− 5x0− 5Δx
= 2x03+ 6x02Δx+ 6x0(Δx )2+ 2(Δx )3− 5x0− 5Δx
The slope of a line through (x0,y0) and (x,y ) is
Δy /Δx= ( y−y0) / (x−x0)
The original function tells us that y0= 2x03− 5x0. In the above equation, let’s substitute
• The quantity [2x03+ 6x02Δx+ 6x0(Δx )2+ 2(Δx )3− 5x0− 5Δx] in place of y
• The quantity (2x03− 5x0) in place of y0
• The quantity (x0+ Δx ) in place of x
When we make these substitutions, we get
Δy /Δx= [2x03+ 6x02Δx+ 6x0(Δx )2+ 2(Δx )3− 5x0− 5Δx− (2x03− 5x0)] / (x0+ Δx−x0)
which can be rewritten as
Δy /Δx= [2x03+ 6x02Δx+ 6x0(Δx )2+ 2(Δx )3− 5x0− 5Δx− 2x03+ 5x0)] / (x0+ Δx−x0)
This can be simplified to
Δy /Δx= [6x02Δx+ 6x0(Δx )2+ 2(Δx )3− 5Δx )] / Δx
and finally to
Δy /Δx= 6x02+ 6x0Δx+ 2(Δx )2− 5
The derivative f ′(x0) is
f ′(x0)= LimΔ →x 0 6x02+ 6x0Δx+ 2(Δx )2− 5
AsΔx approaches 0, the second addend, 6x0Δx, approaches 0 because it is a multiple of Δx. The third ad-
dend, 2(Δx )2, approaches 0 because it’s a constant multiple of (Δx )2. The first addend, 6ax
02, and the fourth
addend,−5, are unaffected by changes in Δx. The limit of the entire expression is therefore 6x02− 5, so f ′(x0)= 6x02− 5
for any real number x0. That means our original function f (x ) = 2x 3− 5x
has the derivative function
f ′(x ) = 6x 2− 5
Here’s another challenge!
In the above scenario, consider the addends in the expression for f as two separate functions g and h, like this:
g (x ) = 2x 3
and
h (x ) = −5x
On this basis, we can say that
f (x ) =g (x ) +h (x )
Show that the derivative of this particular sum is equal to the sum of the derivatives:
f ′(x ) =g′(x ) +h′(x )
Solution
First, let’s figure out the derivative function g′ using the general rule we’ve already derived. Using the rule for the derivative of a variable cubed and then multiplied by a constant, we get
g′(x ) = 6x 2
Using the rule for the derivative of a variable multiplied by a constant, we get
h′(x ) = −5 Adding these results gives us
g′(x ) +h′(x ) = 6x 2− 5
This is what we got when we calculated f ′(x ) in the previous “challenge.”
Are you astute?
After all this repetition, you’ve probably noticed that some of the terms in the examples are always the same as corresponding terms in other examples. You’ve also seen that certain pairs of terms always can- cel out, making the expressions simpler. In the numerators, you always end up subtracting f (x0) from f (x0+ Δx ). In the denominators, you always end up subtracting x0 from x0. In the end, you always get
a ratio of the form
[f (x0+ Δx ) −f (x0)] / Δx
Whenever you calculate a derivative using the techniques in this chapter, you’re actually generating the above expression, and then finding its limit as Δx approaches 0. In fact, the derivative f ′ (x0) of a function f (x ) at the point where x=x0 is commonly defined as
f ′(x0)= LimΔ →x 0 [ f (x0+ Δx ) −f (x0)] / Δx
If we can differentiate the function at every point in its domain, then the formula can be generalized to
f ′(x ) = LimΔ →x 0 [ f (x+ Δx ) −f (x )] / Δx
Remember that Δx can be either positive or negative. The limit, as shown above, must work from both the left and from the right, and the two results must agree. In my opinion, this definition is worth memoriz- ing. You might put it into “mathematical verse,” breaking the thoughts down line by line:
To find the derivative of a function
f of x,
takef of “x plus a little something,” then subtract f of x from that, then divide by the little something,
and finally find the limit as the little something
shrinks
until it’s practically nothing.
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems. Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
1. Find the derivative of the quadratic function: f (x ) =x 2
Here’s a hint for this exercise and all the rest to follow: You may use the general formulas we derived in the text to find these derivatives. Those formulas are theorems now—facts we can use again and again—because we’ve proven them true. You don’t have to go through the ritual of setting up a movable point, approximating the slope, and then finding the limit of that slope as Δx approaches 0.
2. Find the derivative of the quadratic function: f (x ) =x 2
at the points where x= 0, x= 1, x= 2, and x= 3. 3. Find the derivative of the quadratic function:
f (x ) = −2x 2
at the points where x= −3,x= −2, and x= −1. 4. Find the derivative of the quadratic function:
f (x ) = −2x 2
at the points where x= 0, x= 1, x= 2, and x= 3. 5. Find the derivative of the quadratic function:
f (x ) = −7x 2+ 2x
at the points where x= −3,x= −2, and x= −1. 6. Find the derivative of the quadratic function:
f (x ) = −7x 2+ 2x
at the points where x= 0, x= 1, x= 2, and x= 3. 7. Find the derivative of the cubic function:
f (x ) = 5x 3
at the points where x= −3,x= −2, and x= −1. 8. Find the derivative of the cubic function:
f (x ) = 5x 3
at the points where x = 0, x = 1, x = 2, and x = 3.
9. Find the derivative of the cubic function:
f (x ) = 2x 3− 5x
at the points where x = −3,x= −2, and x = −1. 10. Find the derivative of the cubic function:
f (x ) = 2x 3− 5x
In Chap. 3, every function we saw was differentiable, meaning that it had a derivative at every possible point. But there are functions for which derivatives don’t exist at certain points. In this chapter, we’ll learn how to tell when a function is differentiable, and when it is not.
Let’s Look at the Graph
There’s a quick and easy way to see if a function is differentiable at every point in its domain, or if there are some points at which it’s nondifferentiable (impossible to differentiate). We can simply graph it! If the graph appears to have a definable slope at every point, then the function is probably differentiable. We can look for three signs that indicate differentiation problems.
Is there a gap?
Figure 4-1 is a graph of the function f (x)= 1/x. This graph has a gap. We can’t define the function when x= 0. Because the function has no value when x= 0, the graph can’t have a slope at the point where x= 0.
It’s tempting to make an intuitive leap here. Could a line running straight up and down along the axis labeled f (x ) be tangent to the graph? That’s a fine idea, but there’s no point on the graph at which such a line could come into contact! A tangent line always touches a curve at a point.
Because we can’t define the slope of a line tangent to the graph of this function at the point where x= 0, we can’t define the derivative there, either. Even if we add the point (0,0) to definef ( x ) = 1/x for all real numbers, we can’t make the function differentiable at the point where x= 0, as we’ll see later in this chapter.
Is there a jump?
Figure 4-2 is a graph of a function that takes a jump. This function is defined for all real num- bers, although there is a discontinuity. At every point on the line except the one where the jump occurs, the slope, and therefore the derivative, appears to be 0. Let’s look closely at the point where x= 0. At this point, we have f (0) = 3. What is the slope of this graph at (0,3)?
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