At the end of this session you will be able to
THE CHAIN RULE FOR DIFFERENTIATING COMPOSITE FUNCTIONS
When you review the concept of "Composite Functions" on page 19, do notice that g(x) is an independent variable for the function f, but a dependent variable for the function g. This can be confusing because usually the notation g(x) usually refers to dependent variables only. The following are other examples of composite functions:
Example 1: If y = f(u) = sin (u) and u = g(x) = x + 1, then f(g(x)) = sin (x + 1).
Example 2: If y = f(u) = sin (u) and u = g(x) = x + 1, then g(f(u)) = (sin (u) ) + 1.
Example 3: If y = f(u) = tan (u) and u = g(x) = 1 - t, then
Example 4: If y = f(u) = tan (u) and u = g(x) = 1 - t, then
In your textbook, your studying in Section 2.5 on pages 187 and 188, including Examples 1 and 2, are giving you background leading up to the Chain Rule in the box on page 188.
The reason the derivative of the composite function is called the "Chain Rule" is because the quantity "u" is both an independent variable and dependent variable and occurs in both the bottom of dy/du and the numerator of du/dx. Remember that dy/dx, dy/du, and du/dx are NOT fractions though and you cannot be tempted to divide the "u". As a review, recall the meanings of these derivative symbols.
- dy/dx means the derivative of y with respect to x, x being an independent variable.
- dy/du means the derivative of y with respect to u, u being an independent variable here.
- du/dx means the derivative of u with respect to x, being an independent variable.
The discussion after the Chain Rule Theorem 3 on page 188 and at the top of page 189 is sometimes called the "intuitive obvious proof" of the Chain Rule. To understand what the text is saying at the top of page 189, we need to recall that "u" there is independent in delta y / delta u, but dependent in delta u / delta x. Now, if the dependent variable u = g(x) = 12, then delta u = 0 and we would have division by zero in delta 7 / delta u. Therefore, this "intuitive obvious proof" of the Chain Rule would fail. The proof is in Appendix 3 of your textbook.
APPLYING THE CHAIN RULE USING SUBSTITUTION
On page 189 of your textbook, Example 3 "Applying the Chain Rule" illustrates using substitution. In x(t) = cos (t2 + 1), the textbook substituted u for t2 + 1. Then you have cos (t2 + 1) = cos (u) and u = t2 + 1.
- In the solution on page 189, first you subsequently find the derivatives delta x / delta u = - sin (u) and delta u / delta t = 2t.
- After finding the derivatives, the textbook substituted into the chain rule, giving the derivative (- sin (u)) ( 2t ).
- To finish the solution, the textbook substituted u = t2 + 1, giving the final result after simplifying.
Let us look at another example. If y = f(u) = sin u and u = g(x) = x2 - 4, then what is f(g(x))?
1. Find the derivatives delta y / delta u = cos u and delta u / delta x = 2x.
2. Substituting into the chain rule, you have the derivative
3. To finish the solution, you now substitute u = g(x) = x2 - 4 and simplifying, giving the final result
APPLYING THE CHAIN RULE WITHOUT USING SUBSTITUTION
At the bottom of 189, "The Outside-Inside Rule: is applying the Chain Rule without substitution.
Another way to think about the chain rule in the first line of this section is this way: If y = f(g(x)), then you have y = f ( a quantity) and then
the derivative of y with respect to x, delta y / delta x , is f'( a quantity) times the derivative of the quantity.
In Example 4 of page 189 for sin (x2 + x), the "quantity" is x2 + x. Your textbook calls the "quantity" the "inside".
Let us look at the Example 3 on page 189, but applying the Chain Rule without using substitution. What is the derivative of the x(t) = cos (t2 + 1)?
Solution: We select t2 + 1 as our quantity. Then
the derivative of x with respect to t, delta x / delta t , is derivative of cos( a quantity) times the derivative of the quantity.
x'(t) = delta x / delta t = -sin (quantity) times derivative of the quantity
x'(t) = delta x / delta t = -sin (t2 + 1) times (2t)
x'(t) = delta x / delta t = -2t sin (t2 + 1)
Sometimes you need to use the Chain Rule repeatedly. Repeated use of the Chain Rule is illustrated at the top of page 190 in your textbook. Open your book now to page 190 and look at Example 5.
First, the differentiation with the Chain Rule uses 5 - sin (2t) as the "quantity", the "inside", in other words u = 5 -sin (2t).
Second, the differentiation with the Chain Rule uses 2t as the "quantity", the "inside", in other words now u = 2t.
APPLYING THE CHAIN RULE TO POWER FUNCTIONS
In applying the chain rule to power functions, you have nothing new, except that you are using f(u), f( a quantity) is un or (a quantity)n.
For example, the derivative of sin5 x, that is the derivative of (sin x)5, would use sin x as the "quantity", the "inside".
The derivative of sin5 x = derivative of (sin x)5 = 5(sin x)4 times the derivative of (sin x).
Because the derivative of sin x is
we have derivative of sin5 x = 5(sin x)4 times cos x.
Now you can go to the Session 10 assignments to continue your assignment.
At the end of this session you will be able to
DERIVATIVES WITH PARAMETRIC EQUATIONS
If we can eliminate the parameter when you have parametric equations, then you can differentiate as you have learned already in this course. Frequently, it is not easy to eliminate the parameters and, even if you can, you end up with a function that is extremely difficult, though not impossible, to differentiate.
As an example, suppose you have x = 2 sin t, y = 3 cos t, y = f(x) and y > 0. You can use the Pythagorean Identity to eliminate the parameter t, but you will end up with a radical when you solve for y.
Therefore, in your textbook studying you learned how to differentiate without eliminating the parameter. The "Parametric Formula" says that the derivative of a quantity y with respect to x is found by finding the (derivative of the quantity y with respect to the parameter t) is divided by( the derivative of x with respect to the parameter t).
Simply stated, the derivative of a dependent variable with respect to x = division of (derivative of a dependent variable) divided by (the derivative of a independent variable).
In the example above, the quantity y is 2 sin t and the independent variable is x = 3 cos t. The dependent variable quantity is 2 sin t and the independent variable is x = 3 cos t. Therefore,
dy / dx = (dy/dt) / (dx/dt) = 2 cos t / -3 sin t or - (2/3)( cot t).
HIGH ORDER DERIVATIVES WITH PARAMETRIC EQUATIONS
If you want the second order derivative, then the dependent variable quantity y' is what you are differentiating. That is the second order derivative is the derivative of y' with respect x. There using the Parametric Formula from page 190, you have the second order derivative d y'/dx (derivative y' with respect to t) divided by (derivative x with respect to t).
In our example above, because (d y' /dt) = (d (2 cos t)) /dt), the second order derivative is
d y'/dx = (d y' /dt) / (dx/dt) = -2 sin t / -3 sin t or 2/3.
On page 191 of your textbook, the steps in the box in the left margin represent a wonderful guide when finding the second order derivative.
Now you can go to the Session 10 assignments to continue your assignment.