Dr. Gupta gave a couple of very important hints about the upcoming midterm exam for Math 1500 (and she is not one to mislead students).

 

She also said to make sure to know how to do the derivative of an absolute value function.  I discuss this in Lesson 5 (Differentiation Rules), Practise Problem 11 (page 160).

 

The key fact is:

 

( |u| )' = sgn(u) * u '

 

Where "sgn" is short for "signum".  The signum function is merely computing the sign.

 

sgn(u) is saying "the sign of u", i.e. is u negative or positive?  Not your problem.  You just simply say, the derivative of the absolute value of u is sgn(u) * u '.

 

In my example below, x^3 is my designation of x to the power of 3 (x-cubed) and x^2 is saying x to the power of 2 (x-squared):

 

y ' = sgn( x^3 - x) * (3x^2 - 1)