# Unit 3 Chapter 2B (BC Test Date 9/22/15)

Unit 3 – Chapter 2b (BC test date 9/22/15)

- Hmw 2.4a – pg. 137: 9-36(x3), 37, 38, 43, 44 (Chain rule)
- Hmw 2.4b – pg. 137: 45-97 every other odd (Chain rule)
- Hmw 2.4c – pp. 139-140: 109, 110, 113, 115, 120, 123, 131 (Chain rule)
- Hmw 2.5a – pp. 146: 1-37 EOO (Implicit differentiation)
- Hmw 2.5b – pp. 147-148: 41, 45, 48, 51, 53, 57, 59, 65, 73, 74, 77 (Implicit differentiation)
- Hmw 2.6a – pp. 154: 1-4, 5 & 7 IN CLASS TOGETHER AFTER THE LECTURE (Related Rates)
- Hmw 2.6b – pp. 154-155: 11 – 25 odd
- Hmw 2.6c – pp. 155-157: 31, 33, 43, 44, 50, 53

Objectives – Students will be able to…

- Find derivatives using the Chain Rule and the General Power Rule.
- Simplify derivatives using algebra.
- Use the Chain Rule to find the derivative of trigonometric functions.
- Distinguish between functions written in implicit form and explicit form.
- Use implicit differentiation to find the derivative of a function.
- Find related rates and use them to solve real-life problems.

Chain Rule

If y = f(g(x)) then .

Sometimes teachers refer to the chain rule as the “peanut m&m” rule.

A function like this one… … is like a peanut m&m. The ( )4 is like the shell, and the 3x + 2 is like the peanut.

To find the derivative of this kind of function, find the derivative of the shell and multiply it to the derivative of the peanut.

m&m (The function) / Derivative of the shell / Derivative of the peanut / Final answer:/ / 3 /

/ / 8x3 /

The chain can be applied multiple times, as many as needed to completely unravel the chain.

Ex: think of this as . So, ( )4 is the outer shell, sin( ) is the “chocolate coating”, and 3x+4 is the peanut. To find the derivative you need to find d(shell)/dx, d(chocolate)/dx, and d(peanut)/dx. Then multiply.

**Use the Chain Rule to find the Derivative.**

**Note: #5 and #8 cannot be done because their radicands will always be negative, and even indexed roots of negative radicands are not real numbers.**

**Implicit Differentiation**

An Explicit equation is one where y is explicitly written as a function of x.

Ex: y = 3x2+ 2

So far you have only found derivatives for explicit functions.

An Implicit equation is one that, basically, isn’t solved explicitly for y.

Ex: x2 + y3 = 6

To find the derivative of an implicit equation. Take the derivative of both sides of the equation and treat y as a sort of “unknown” function. Since it is “unknown” the only thing we can write for its derivative is dy/dx.

Ex: Given x2 + y3 = 6 applying implicit differentiation gives .

Notice how y3 was treated like a chain rule problem. If we had (f(x))3, chain rule would make the derivative 3(f(x))2f’(x). Here it is the same idea. The y3 is the shell and y is the peanut. The d(shell)/dx is 3y2. The d(peanut)/dx is dy/dx.

To finish the example, solve for dy/dx. So the derivative of x2 + y3 = 6 by implicit differentiation is…

Related Rates

Chain rule and implicit differentiation work together to solve related rates problems. The textbook gives the following tips for solving these problems.

- Identify all given quantities and quantities to be determined. Make a sketch and label the quantities. (This is a very important step. Make an actual list of these “known” and “to be determined” things.)
- Write an equation involving the variables whose rates of change either are given or are to be determined. (Almost “step 1A”. You want to make a list of the dy/dt or dr/dt or whatever rates you actually know and whatever rates you need to find. And then write an equation using the variables and rates from step 1 and “1A”.)
- Use the chain rule and implicit differentiation to find the derivatives of both sides of the equation with respect to time.
- Plug in all of the “knowns” into your derivative and solve for the unknown rate.