We have just recreated an example of the original “Newton-Leibniz magic,” which allows us to find the slope of a line tangent to a curve, or to find the instantaneous rate of change in a nonlinear function, at a specified point. Now let’s work out some derivatives over whole domains for some basic linear functions.
Simply a constant
Imagine a constant function in which the dependent variable always has the same value, no matter what we input for the independent variable. If we call the function f, then we can describe it simply enough:
f (x ) =a
where x is the independent variable and a is a real number. Let’s find the derivative function f ′ at some unspecified point (x0,y0).
We begin by creating a movable point (x,y ) near (x0,y0), as shown in Fig. 3-5. In the rect-
angularxy-plane,y=a for all possible values of x. The x-value of our movable point is x0+ Δx.
They-value is a. The derivative at (x0,y0) can be approximated by calculating the slope of a
straight line passing through (x0,y0) and (x,y ). That slope is
Δy /Δx= ( y−y0) / (x−x0)
If we substitute a for y, a for y0, and the quantity (x0+ Δx ) for x in this equation, we get
Δy /Δx= (a−a) / (x0+ Δx−x0)
which can be simplified to
Δy /Δx= 0/Δx= 0 Slope of line = Δy/Δx f(x) = a (x0,y0) Movable point (x,y) = (x0+Δx,y0+Δy) Δx Δy= 0
Figure 3-5 Finding the derivative of the constant functionf (x)=a at a point (x0,y0).
Δy Δx Slope of line = Δy/Δx f(x) = ax (x0,y0) (x,y) = Movable point (x0+Δx,y0+Δy)
Figure 3-6 Finding the derivative of the basic linear function
f (x)=ax at a point (x0,y0).
The derivative f ′(x0) is
f ′(x0)= Lim
x
Δ →0 0
No limit can be more straightforward than this! The derivative f ′(x0) is equal to 0, no matter
what we choose for x0. By doing this exercise, we’ve shown that for any constant function
f (x ) =a the derivative function is
f ′(x ) = 0 which is called, appropriately enough, the zero function.
Multiplyx by a real constant
Now we’ll find the derivative of a basic linear function in which the independent variable is multiplied by a real-number constant. Again, let’s call the function f. Then
f (x ) =ax
where x is the independent variable and a is the constant. Let’s figure out f ′ at a nonspecific point (x0,y0).
We can invent a movable point (x,y ) near (x0,y0), as shown in Fig. 3-6. In rectangular
xy-coordinates, y=ax for all possible values of x and y. The x-value of our movable point is x=x0+ Δx
They-value of the movable point is
y=ax=a(x0+Δx ) =ax0+aΔx
We can approximate the derivative by finding the slope of a straight line passing through the points (x0,y0) and (x,y ). That slope is
Δy /Δx= ( y−y0) / (x−x0)
We know that y0=ax0, because f tells us that any x-value must be multiplied by a to get the
corresponding y-value. In the above equation, let’s substitute the quantity (ax0+aΔx ) for y,
substituteax0 for y0, and substitute the quantity (x0+ Δx ) for x. That gives us
Δy /Δx= (ax0+aΔx−ax0) / (x0+ Δx−x0)
which can be simplified to
Δy /Δx=a(x0+ Δx−x0) / Δx
and further to
Δy /Δx=aΔx /Δx and finally to
Δy /Δx=a The derivative f ′(x0) is therefore
f ′(x0)= Δ →Limx 0 a
This limit here is equal to a. That’s all there is to it! Therefore, f ′(x0)=a
no matter what value of x0 we choose. This result tells us that for any basic linear function of
the form
f (x ) =ax the derivative function is
f ′(x ) =a which is a constant function.
Here’s a challenge!
Using the calculus techniques we’ve learned so far (not plain algebra!), find the derivative of the function
f (x ) = −4x+ 5
at the general point (x0,y0) in rectangular xy-coordinates, where y=f (x ). What does this result tell us? Solution
Once again, let’s start out by creating a movable point (x,y ) near (x0,y0). Figure 3-7 shows the slant of the
line. (The coordinate grid is not shown because it would make the illustration needlessly messy. It’s the slope we’re interested in, anyhow!). In rectangular xy-coordinates, we can graph our function in the slope- intercept form as
y= −4x+ 5 Thex-value of our movable point is, as always,
x=x0+ Δx
They-value of our movable point is
y= −4(x0+ Δx ) + 5 = −4x0− 4Δx+ 5 Δy Δx Slope of line = Δy/Δx (x0,y0) Movable point (x,y) = (x0+Δx,y0+Δy) f(x) = –4x+ 5 Figure 3-7 Finding the derivative of the
functionf (x) = −4x+ 5 at a point (x0,y0).
We approximate the derivative by finding the slope of a straight line through (x0,y0) and (x,y ). That slope is
Δy /Δx= ( y−y0) / (x−x0)
Our function f tells us that y0= −4x0+ 5. Now, in the above equation, let’s substitute the quantity (−4x0−
4Δx+ 5) in place of y, substitute the quantity (−4x0+ 5) in place of y0, and substitute the quantity (x0+ Δx )
in place of x. That gives us
Δy /Δx= [−4x0− 4Δx+ 5 − (−4x0+ 5)] / (x0+ Δx−x0)
We had better be careful with the signs here! We can rewrite the above equation as Δy /Δx= (−4x0− 4Δx+ 5 + 4x0− 5) / (x0+ Δx−x0)
which can be simplified to
Δy /Δx= −4Δx /Δx
and finally to
Δy /Δx= −4
We got rid of a big mess in a hurry! Now we can say that the derivative f ′(x0) is f ′(x0)= Lim
x
Δ →0 −4 This is, again, trivial. The limit is equal to −4. Therefore,
f ′(x0)= −4
no matter what value of x0 we choose. This tells us that the derivative of the linear function f (x ) = −4x+ 5
is the function
f ′(x ) = −4
The stand-alone constant, 5, in f (x ) can be changed to any other real number, and we’ll still end up with the same derivative function, f ′(x ) = −4.