Programmed tutorial: Derivative of the Absolute Value of a Linear Expression

Find the derivative:

Do we treat this with the same rule as for
?

No.

State the rule we must apply in terms of a general function u.

What effect does the factor have?

It generates the correct algebraic sign.

What is the expression for u here?

u = 3x – 7

What is the value of

Substitute these results in the rule for the derivative above.

Will this always work out to a value of 3?

No.

Which factor in
can cause the expression to be negative?

Since the denominator is an absolute value, and is always positive, we need to look only at the (3x – 7) factor in the numerator.

Find the values of x which make (3x – 7) negative.

We set (3x – 7) < 0.

Solve for x.

We get 3x < 7, or x < 7/3 in order for the derivative of |3x – 7| to be negative.

Check this for x = 1, which is certainly less than 7/3.

For x = 1,

So we see that whenever the value of x is less than 7/3, the derivative will be –3.

To get more of a feel for this problem, let’s plot it in stages.  First plot a related function, y = 3x – 7 (no absolute value), on paper.  Then check your graph by clicking on “Next”.

alt="Graph of a linear function Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

What do we call the marked point?

It is the x-intercept. What is the value of y there?

Since the point is on the x-axis, its y-coordinate is 0.

Is the slope of this graph the same everywhere?

Yes, it is +3.

Is the y-coordinate positive everywhere?

No.

Where is it negative?

For x-values to the left of the x-intercept, the line is below the x-axis and the y-coordinate is negative.  That is the red portion of the line.

Next we will plot y = |3x – 7|.  Will any part of the graph will be the same as the first graph?

Yes.

Describe the part that is the same.

Where 3x – 7 is positive, |3x – 7| = 3x – 7, and the graphs will be the same.

For what values of x will this be true?

For values to the right of the intercept, or

Those would be x-values along the blue part of the x-axis.

How do we need to change the part of the graph to the left of the x-intercept (for x-values along the green part of the x-axis)?

Since 3x – 7 is negative there, we have |3x – 7| = – (3x – 7) for

What do we need to change on the previous graph to plot y = |3x – 7|?

We need to multiply any negative y-coordinate by –1.

For example, how would we change the point
at (1, – 4) in the previous graph?

We would plot it at (1, 4).

Will this new point be above or below the x-axis?

Above the x-axis.

Will all of the points on the previous graph that were below the x-axis be above the axis in our plot
of y = |3x – 7|?

Yes.

Now plot y = |3x – 7| on paper.  Then check your graph by clicking on “Next”.

alt=" Graph of the absolute value of a linear expression Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!


We see that the black part of the first graph is unchanged.  The red part of the first graph is reflected across the x-axis so that we have positive values of the y-coordinate.  This is consistent with y = |3x – 7| always being positive.
Notice that the slope is negative for x to the left of the x-intercept.

Recall that we found that 

When x is to the right of the x-intercept (x is on the blue part of the axis), find the sign of the

factor in the derivative.

Since 3x – 7 is positive for this region, we can replace the absolute value of 3x – 7 by itself:
We have
(for x-values in the blue region).  Hence the slope is +3, as we see from the diagram for the black part of the line.

Now, when x is to the left of the x-intercept (x is on the green part of the axis), find the sign of the 

factor in the derivative.

Since 3x – 7 is negative for this region, we can replace the absolute value of 3x – 7 by its negative:
We have

Hence the slope is – 3, as we can also see from the diagram.

Summarize these results.

Our application of the Absolute Value Rule gave us

In other words, the derivative of the absolute value is the product of a “sign factor” and the derivative of the “stuff” between the absolute value signs.
The “sign factor” is +1 or – 1.

The end. If you found this helpful and would recommend that I create more pages like this one, please let me know: Email to John Taylor