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# Kalkulus II (25 - 26)

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### Kalkulus II (25 - 26)

1. 1. Kalkulus II<br />Teguh Budi P, M.Si <br />Sesion#25-26<br />JurusanFisika<br />FakultasMatematikadanIlmuPengetahuanAlam<br />
2. 2. Partial Derivatives<br />The Chainrule<br />2<br />Outline<br />1/9/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
3. 3. Multivariable Functions and Their Derivatives(part 3)<br />3<br />1/9/2011<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
4. 4. z<br />100<br />An Introduction to Partial Derivatives<br />10<br />y<br />10<br />x<br />1/9/2011<br />4<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
5. 5. When we have functions with more than one variable, we can find partial derivatives by holding all the variables but one constant.<br />z<br />100<br />10<br />y<br />10<br />x<br />(eff sub ecks)<br />Note:<br />is also written as<br />1/9/2011<br />5<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
6. 6. z<br />would give you the slope of the tangent in the plane y=0 or in any plane with constant y.<br />100<br />10<br />y<br />10<br />x<br />In other words, how is changing one variable going to change the value of the function?<br />1/9/2011<br />6<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
7. 7. Mixed variables are also possible:<br />Both answers are the same!<br />1/9/2011<br />7<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
8. 8. The Chain Rule<br />Consider a simple composite function:<br />1/9/2011<br />8<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
9. 9. and another:<br />1/9/2011<br />9<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
10. 10. This pattern is called the chain rule.<br />and one more:<br />1/9/2011<br />10<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
11. 11. If is the composite of and , then:<br />Find:<br />example:<br />Chain Rule:<br />1/9/2011<br />11<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
12. 12. We could also do it this way:<br />1/9/2011<br />12<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
13. 13. Here is a faster way to find the derivative:<br />Differentiate the outside function...<br />…then the inside function<br />1/9/2011<br />13<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
14. 14. Another example:<br />It looks like we need to use the chain rule again!<br />derivative of the<br />outside function<br />derivative of the<br />inside function<br />1/9/2011<br />14<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
15. 15. The chain rule can be used more than once.<br />Another example:<br />(That’s what makes the “chain” in the “chain rule”!)<br />1/9/2011<br />15<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
16. 16. The formulas on the memorization sheet are written with instead of . Don’t forget to include the term!<br />Derivative formulas include the chain rule!<br />etcetera…<br />1/9/2011<br />16<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
17. 17. The most common mistake on the chapter 3 test is to forget to use the chain rule.<br />Every derivative problem could be thought of as a chain-rule problem:<br />The derivative of x is one.<br />derivative of outside function<br />derivative of inside function<br />1/9/2011<br />17<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
18. 18. Divide both sides by<br />The slope of a parametrized curve is given by:<br />The chain rule enables us to find the slope of parametrically defined curves:<br />1/9/2011<br />18<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
19. 19. Example:<br />These are the equations for an ellipse.<br />1/9/2011<br />19<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />
20. 20. Thank You<br />1/9/2011<br />20<br />© 2010 Universitas Negeri Jakarta | www.unj.ac.id |<br />