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# Chain Rule

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### Chain Rule

1. 1. THE CHAIN RULE Chapter 3 Section 6
2. 2. The Purpose of Chain Rule <ul><li>The chain rule is an effective yet easier way of finding the derivative of more complicated equations. </li></ul><ul><li>Look at this problem: y = (x 3 + 5x) 7 </li></ul>
3. 3. Chain Rule Contd. <ul><li>y = (x 3 + 5x) 7 </li></ul><ul><li>Formula to find derivative of above problem: </li></ul><ul><li>y = (u) 7 ; u= x 3 + 5x : get derivative of “u”, which is: </li></ul><ul><li>3x 2 + 5 = du : now, to find y’, </li></ul><ul><li> y’= 7u 7-1 x du or 7u 6 x du : plug in “u” and “du” to get: </li></ul><ul><li>7(x 3 + 5x) 6 (3x 2 + 5) : distribute the 7 and multiply </li></ul><ul><li>y’= (21x 2 + 35)(x 3 + 5x) </li></ul>
4. 4. The Chain Rule & Trig. Functions <ul><li>Find the derivative using chain rule: </li></ul><ul><li>y = sin 3 2x find y’ </li></ul><ul><li>1: sin 3 2x = (sin 2x) 3 [changing makes it easier] </li></ul><ul><li>our “u” is: (sin 2x) </li></ul><ul><li>our “du” is: 2cos 2x </li></ul><ul><li> y= u 3 x du </li></ul><ul><li> y’= 3u 2 x du ==> plug in “u” and “du” into equation. </li></ul><ul><li> y’= 3(sin 2x) 2 (2 x cos 2x) </li></ul><ul><li> y’= 6(sin 2x) 2 (cos 2x) </li></ul>
5. 5. Using Inside-Outside Method <ul><li>The following problem can be solved 2 ways: </li></ul><ul><li>y= sin(3x + 1) </li></ul><ul><li>-The first way we’ll do is the chain rule: </li></ul><ul><li>“ u” = 3x + 1 making our “du” = 3 </li></ul><ul><li>y= sin u x du y’= cos u x du ==> plug in “u” and “du” </li></ul><ul><li>y’= cos (3x +1) x 3 = 3 cos (3x + 1) </li></ul>
6. 6. Using Inside-Outside Method contd. <ul><li>y= sin(3x + 1) </li></ul><ul><li>To solve using Inside-Outside Method: </li></ul><ul><li>Take the derivative of the outside of the equation (sin(3x +1)) </li></ul><ul><li>Take the derivative of the inside of the equation (3x +1) </li></ul><ul><li>Derivative of outside (sin (3x +1)) = cos(3x +1) </li></ul><ul><li>Derivative of inside (3x +1) = 3 </li></ul><ul><li>Put them together to GET: </li></ul><ul><li>y’= 3 cos(3x + 1) </li></ul>
7. 7. Chain rule and Radicals Solve for y’ using the Chain rule To SOLVE: = (9x 2 +4) 1/2 “u” = 9x 2 +4 “du” = 18x y = u 1/2 x du y’= 1/2u -1/2 x du [plug in “u” and “du”] y’= 1/2(9x 2 + 4) -1/2 x 18x : multiply 18x by 1/2 to get: 9x (9x 2 + 4) -1/2 =
8. 8. Chain rule and Radicals Find y’ To solve, you will use the chain rule and inside outside rule. = (tan(3x)) 1/2 u= tan (3x) To find du, use inside outside rule. So, u = tan (3x) Derivative of tan (3x) [outside: sec 2 (3x)] Derivative of (3x) [inside: 3] du = 3 tan(3x) continue on next slide…
9. 9. Chain rule and Radicals <ul><li>u= tan (3x) and du= 3 tan (3x) </li></ul><ul><li>y= u 1/2 du | y’=1/2u -1/2 x du [plug in “u” and “du”] </li></ul><ul><li>y’= 1/2(tan 3x) -1/2 x 3 tan 3x </li></ul><ul><li> </li></ul><ul><li>= </li></ul>
10. 10. THE END!