Chapter 06

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Chapter 06

  1. 1. 1Chapter 6Differential CalculusThe two basic forms of calculus aredifferential calculus and integral calculus.This chapter will be devoted to the formerand Chapter 7 will be devoted to thelatter. Finally, Chapter 8 will be devotedto a study of how MATLAB can be used forcalculus operations.
  2. 2. 2Differentiation and the DerivativeThe study of calculus usually begins withthe basic definition of a derivative. Aderivative is obtained through the processof differentiation, and the study of all formsof differentiation is collectively referred toas differential calculus.If we begin with afunction and determine its derivative, wearrive at a new function called the firstderivative. If we differentiate the firstderivative, we arrive at a new functioncalled the second derivative, and so on.
  3. 3. 3The derivative of a function is theslope at a given point.
  4. 4. 4Various Symbols for the Derivative( )or ( ) ordy df xf xdx dx0Definition: limxdy ydx x∆ →∆=∆
  5. 5. 5Figure 6-2(a). Piecewise LinearFunction (Continuous).
  6. 6. 6Figure 6-2(b). Piecewise LinearFunction (Finite Discontinuities).
  7. 7. 7Piecewise Linear Segment
  8. 8. 8Slope of a Piecewise Linear Segment2 12 1slopey ydydx x x−= =−
  9. 9. 9Example 6-1. Plot the first derivativeof the function shown below.
  10. 10. 10
  11. 11. 11Development of a Simple Derivative2y x=2( )y y x x+ ∆ = + ∆2 22 ( )y y x x x x+ ∆ = + ∆ + ∆
  12. 12. 12Development of a Simple DerivativeContinuation22 ( )y x x x∆ = ∆ + ∆2yx xx∆= + ∆∆0lim 2xdy yxdx x∆ →∆= =∆
  13. 13. 13Chain Rule( )y f u= ( )u u x=( )( )dy df u du duf udx du dx dx= =( )( )df uf udu=where
  14. 14. 14Example 6-2. Approximate the derivativeof y=x2at x=1 by forming small changes.2(1) (1) 1y = =2(1.01) (1.01) 1.0201y = =1.0201 1 0.0201y∆ = − =0.02012.010.01dy ydx x∆≈ = =∆
  15. 15. 15Example 6-3. The derivative of sin uwith respect to u is given below.( )sin cosdu udu=Use the chain rule to find thederivative with respect to x of24siny x=
  16. 16. 16Example 6-3. Continuation.2u x=2duxdx=2( )4(cos )(2 ) 8 cosdy du dy duf udx dx du dxu x x x= == =
  17. 17. 17Table 6-1. Derivatives( )f x ( )f x Derivative Number( )af x ( )af x D-1( ) ( )u x v x+ ( ) ( )u x v x+ D-2( )f u ( )( )du df u duf udx du dx=D-3a 0 D-4( 0)nx n ≠ 1nnx − D-5( 0)nu n ≠ 1n dunudx−D-6uv dv duu vdx dx+ D-7uv2du dvv udx dxv− D-8ue u duedxD-9
  18. 18. 18Table 6-1. Derivatives (Continued)ua( )ln u dua adxD-10lnu 1 duu dxD-11loga u( )1logadueu dxD-12sinucosduudx   D-13cosusinduudx− D-14tanu 2secduudxD-151sin u−121sin2 21duudxuπ π− − ≤ ≤  −D-161cos u−211dudxu−−( )10 cos u π−≤ ≤ D-171tan u−121tan1 2 2duuu dxπ π− − < < +  D-18
  19. 19. 19Example 6-4. Determine dy/dx forthe function shown below.2siny x x=( ) ( )22 sinsindy dv duu vdx dx dxd xd xx xdx dx= += +
  20. 20. 20Example 6-4. Continuation.( )22cos sin 2cos 2 sindyx x x xdxx x x x= += +
  21. 21. 21Example 6-5. Determine dy/dx forthe function shown below.sin xyx=( ) ( )2 22sinsincos sind x d xdu dvv u x xdy dx dx dx dxdx v xx x xx− −= =−=
  22. 22. 22Example 6-6. Determine dy/dx forthe function shown below.22xy e−=22xu = −( )22 122xddux xdx dx − ÷  = = − = − ÷ ( )2 22 2x xdye x xedx− −= − = −
  23. 23. 23Higher-Order Derivatives( )y f x=( )( )dy df xf xdx dx= =2 22 2( )( )d y d f x d dyf xdx dx dx dx = = =  ÷ 3 3 2(3)3 3 2( )( )d y d f x d d yf xdx dx dx dx = = =  ÷ 
  24. 24. 24Example 6-7. Determine the 2ndderivative with respect to x of thefunction below.5sin 4y x=5(cos4 ) (4 ) 20cos4dy dx x xdx dx= × =( )2220 sin 4 (4 ) 80sin 4d y dx x xdx dx= − × = −
  25. 25. 25Applications: Maxima and Minima1. Determine the derivative.2. Set the derivative to 0 and solve forvalues that satisfy the equation.3. Determine the second derivative.(a) If second derivative > 0, point is aminimum.(b) If second derivative < 0, point is amaximum.
  26. 26. 26DisplacementVelocityAccelerationdyvdt=Displacement, Velocity, and Accelerationy22dv d yadt dt= =
  27. 27. 27Example 6-8. Determine local maximaor minima of function below.3 2( ) 6 9 2y f x x x x= = − + +23 12 9dyx xdx= − +23 12 9 0x x− + =1 and 3x x= =
  28. 28. 28Example 6-8. Continuation.23 12 9dyx xdx= − +For x = 1, f”(1) = -6. Point is a maximum andymax= 6.For x = 3, f”(3) = 6. Point is a minimum andymin = 2.226 12d yxdx= −

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