ABSOLUTE VALUE DERIVATIVES
Find the derivative of:
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.
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.
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.
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.
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.