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…

  1. Find derivatives using the Chain Rule and the General Power Rule.
  2. Simplify derivatives using algebra.
  3. Use the Chain Rule to find the derivative of trigonometric functions.
  4. Distinguish between functions written in implicit form and explicit form.
  5. Use implicit differentiation to find the derivative of a function.
  6. 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))2f’(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.

  1. 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.)
  2. 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”.)
  3. Use the chain rule and implicit differentiation to find the derivatives of both sides of the equation with respect to time.
  4. Plug in all of the “knowns” into your derivative and solve for the unknown rate.