1. DIFFERENTIATION Laura Berglind Avery Gibson

2. Definition of Derivative: Lim f(x+h) – f(x) h  0 h Derivative= slope Lim= limit h  0 = as h approaches zero

3. Notations for Derivative f’(x)F prime DyDerivative of y in dxrespect to x y’ y prime

4. Let’s Put the Definition to Practice  Example: f(x)=x ²  We know the answer is: f’(x)= 2x Lim f(x+h) – f(x)=lim x ²+2xh=h²-x²=lim h(2x+h) h  0 hh  0 hh  0 h  When bottom H goes away, you can plug in 0 for h Lim(2x+0) = 2x h  0

5. Thus, you can follow these three simple steps! 1. Replace x with x+h and copy problem 2. Factor and get rid of h 3. Replace h with 0

6. Let’s try one more! Lim f(x+h) – f(x)=lim (x +h)^1/2-x^1/2 h  0 hh  0 h f(x)= √x Think of this problem as: f(x)= x^1/2 lim (x +h)^1/2-x^1/2 * (x +h)^1/2+x^1/2 h  0 h *multiple by the conjugate (x +h)^1/2=x^1/2 lim x +h-x______= lim ___1_______ h  0 h [(x+h)^1/2 + x^1/2]h  0 (x+h)^1/2 + x^1/2 = 1______= 1____=1/2x^-1/2 (x+0)^1/2 +x^1/2 2x^1/2

7. Old Rules to remember! Lim sinh = 1 h  0 h Lim 1-cosh = 0 h  0 h *Sin(x+h) = sinxcosh + cosxsinh *Cos(x+h) = cosxcosh - sinxsinh Before we move on…..

8. Trigonometry  f(x)=sinxf’(x)=cosx  Simple steps, we learned earlier: 1. Replace x with x+h and copy problem 2. Factor and get rid of h 3. Replace h with 0  Now we are going to use these steps to prove f(x)=sinx f’(x)=cosx

9. F(x)=sinxf’(x)=cosx Lim f(x+h) – f(x)=Lim sin(x+h)-sinx h  0 hh  0h = Lim sinxcosh + cosxsinh- sinx=Lim sinx(cosh-1) + Lim cosxsinh h  0 hh  0 h h  0 h = Lim sinx (0) + Lim cosx (1)=Lim cosx h  0 h  0h  0

10. Now let’s prove that f(x)=cosxf’(x)=-sinx Lim f(x+h) – f(x)=Lim cos(x+h)-cosx h  0 hh  0 h = Lim cosxcosh - sinxsinh- cosx=Lim cosx(cosh-1) - Lim sinxsinh h  0 h h  0 h h  0 h =Lim cosx (0) – Lim sinx(1)= -sinx h  0 h  0

11. Now let’s prove that f(x)=tanxf’(x)=sec ²x *Think of f(x)=tanx as f(x)= sinx/cosx Lim = sin(x+h) – sinx H  0 cos(x+h) cosx h *now you need to find a common denominator and then flip! Lim sin(x+h) cosx – sinxcos(x+h)* _1_ H  0 _____cos (x+h) cosx_______ h __h__ 1

12. Tangent continued… Lim sin(x+h) cosx – sinxcos(x+h) H  0 h[cos(x+h) cosx] Lim (sinxcosh + cosxsinh) * cosx- sinx (coscosh – sinxsinh) H  0 hcos(x+h) * cosx Lim sinxcoshcosx + cos ²xsinh – cosxcoshsinx + sin²xsinh *cancel out H  0 h cos (x+h) cosx

13. Tangent continued… Lim sinh (sin ²x + cos²x)=Lim 1 * 1_____ h  0 h cos(x+h) cosxh  0 cos(x+h)cosx *side note: Never expand the bottom!!! Lim 1______=Lim 1___ h  0 cos(x+0)cosxh  0 cos²x *which equals sec²x!!!

14. Memorize these trig functions and their derivatives!  F(x)=sinx F’(x)=cosx  F(x)=cox F’(x)=-sinx  F(x)=tanx F’(x)=sec ²x  F(x)=cotx F’(x)=-csc ²x  F(x)=secx F’(x)=secxtanx  F(x)=cscx F’(x)=-cscxcotx *you can remember all the C’s have negative derivatives!

15. SHORT CUTS!!! 1. Chain Rule 2. Product Rule 3. Quotient Rule

16. Chain Rule  F(x)=u ^ nF’(x)= nu ^ (n-1)  The exponent goes out front and then subtract 1 from the exponent *used every time although may be embedded in product or quotient

17. Example of Chain Rule  f(x)= x ³+6x *bring the 3 and multiple it by the one is front of the x. then subtract 1 from 3 (the exponent). *bring the 1 down to multiple it by the sixth. Then subtract 1 from the exponent, which in this case is zero  f’(x)=3x ² +6x ºThus, f’(x)=3x²+6 *side note: derivative of any constant is 0 For example: f(x)=5f’(x)=0

18. Try Me!!! f(x)=4x ³+10x²-6x+100

19. And the answer is… f(x)= 4x ³+10x²-6x+100 f’(x)=12x ²+20x-6+0  Thus, f’(x)=12x²+20x-6

20. Product Rule  Used when you are multiplying  Equation: (F)(DS)+(S)(DF)  F=first number or set of numbers  DS=Derivative of second number  S=second number or set of numbers  DF=Derivative of first number  For example: (x+2)(5x+2) first second

21. Example of Product Rule 1. f(x)= (2x+1)^5 (5x-3) ² 2. f’(x)=(2x+1)^5(2)(5x-3)(5) + (5x-3)²(5)(2x+1)^4 (2) 3. F’(x)= 10(2x+1)^4(5x-3)[(2x+1)+(5x-3)] 4. F’(x)= 10(2x+1)^4 (5x-3) (7x-2)

22. Try Me!!! f(x) = (4x+1) ²(1-x)³

23. And the Answer is…  f(x) = (4x+1) ²(1-x)³  f’(x)=(4x+1) ²(3)(1-x)²(-1) + (1-x)³(2)(4x+1)(4)  f’(x)= (4x+1)(1-x)² [-3(4x+1)+8(1-x)]  f’(x)= (4x+1)(1-x)²[-12x-3+8-8x]  f’(x)= (4x+1)(1-x)²[-20x+5]  f’(x)= 5(4x+1)(1-x)²(-4x+1)

24. Quotient Rule  Used when you are dividing.  Equation: (B)(DT)-(T)(DB) B ²  B= the bottom number(the denominator)  DT= the derivative of the top number  T= the top number (the numerator)  DB= the derivative of the bottom number  For example: 2+1x Top Number 5x-e Bottom Number

25. Example of Quotient Rule  Y= (2-x)/(3x+1)  Y’= (3x+1)(-1)-(2-x)(3) (3x+1) ²  Y’=(-3x-1)=(-6=3x) (3x+1)²  Y’=___-7___ (3x+1)²

26. Try Me!!! Y= 1+x ² 1-x²

27. And the Answer is… Y= 1+x ² 1-x²  Y’= (1-x ²)(2x) – (1+ x ²)(-2x) (1- x ²)²  Y’= 2x-2x³+2x+2x³ (1- x ²)² Y’= 4x___ (1- x ²)²

28. Natural Log  When taking the derivative of natural log: 1 over the angle * the derivative of the angle  For example: F(x)= ln(x ²+3x) F’(x)= 1__ * (2x+3) x ²+3x

29. Try Me Real Quick! f(x)=ln (x ³+5x²+6x)

30. And the Answer is…  f(x)=ln (x ³+5x²+6x)  f’(x)= 1 (3x²+10x+6) x ³+5x²+6x

31. Derivative of e  f(x)=e f’(x)=e  f(x)=e^2xf’(x)=2e^2x  f(x)=e^10x f’(x)=10e^10x

32. When taking the derivative of an exponent *Copy function and then multiply by the derivative of exponent For example:  F(x)=e ^sinx  F’(x)=e^sinx (cosx)

33. ETA  For only power of trig functions  E= Exponent  T= Trig  A= Angle  Copy the rest of problem! For Example: F(x)=sin^5(cosx) F’(x)= (5)sin^4(cosx)*cos(cosx)*(-sinx)

34. Let’s try one more…  F(x)= tan ²x(5x²-6x+1)  F’(x)=(2)tan(5x²-6x+1)*sec²(5x²-6x+1)*(10x-6) *copy rest of problem after derivative of exponent (chain rule) *copy problem after take derivative of trig

35. Let’s Try a REAl AP Multiple Choice quesiton  If f(x)=sin(e^-x), then f’(x)=? a)-cos(e^-x) b)cox(e^-x)+(e^-x) c)cos(e^-x)-(e^-x) c)(e^-x) cos(e^-x) d)-(e^-x)cos(e^-x)

36. If f(x)=sin(e^-x), then f’(x)=? ANSWER:d)-(e^-x)cos(e^-x) E1. sin^0(e^-x)  1 T2.cos(e^-x) A3.(e^-x)(-1)

37. Let’s Review real quick  y=(4x+1) ²(1-x)³  Y=2-x 3x+1  Y=3x^(2/3)-4x^(½)-2 Try these!

38. y=(4x+1) ²(1-x)³  Y’= (4x+1)²(3)(1-x)²(-1)+(1-x)³(2)(4x+1)(4)  Y’=-3(4x+1) ²(1-x)²+8(1-x)³(4x+1)  Y’=(4x+1)(1-x)²[-3(4x+1)+8(1-x)]  Y’=(4x+1)(1-x)²[-12x-3+8-8x  y’=(4x+1)(1-x)² 5(-4x+1)

39. Y=2-x 3x+1  Y’=(3x+1)(-1)-(2-x)(3) (3x+1) ²  Y’=-3x-1-6+3x (3x+1)²  y’= -7____ (3x+1) ²

40. Y=3x^(2/3)-4x^(½)-2  Y’=(2/3)(3)(x^-1/3)(1)-(4)(1/2)(x^-1/2)(1)-0  Y’=2x^-1/3-2x^-1/2  Y’=2(x^-1/3 – x^-1/2)

41. Implicit Differentiation  Used when x and y are in the problem and they can’t be separated.  When taking the derivative of y... You still do it in respect to x... Everything is in terms of x  Dy dx  always there when taking derivative of y *solve for dy/dx

42. Let’s Try one F(x)= X ²+y²=25 1. 2x+2y(dy/dx) = 0 *take derivative like normal, except whenever you take the derivative of y, you must add in dy/dx 2. *Now, solve for dy/dx 2y(dy/dx)=-2x 3. dy= -2x= -x_ dx 2y y

43. Your Turn!  F(x)=(2x+1)^5+3y²=6

44. F(x)=(2x+1)^5+3y²=6  F’(x)=(5)(2x+4)^4(2)+6y(dy/dx)=0  6y(dy/dx)=-10(2x+4)^4  dy= -10(2x+1)^4 dx 6y  dy= -5(2x+1)^4 dx 3y  *side note: if the dy/dx were to cancel out, that means that the derivative does not exist!

45. Logarithmic Differentiation  Done when there is a variable in the exponent  3 rules to remember 1. Multiplication  addition 2. Division  subtraction 3. Exponents  multipliers

46. For example: y=2^x  Lny= xln2 *RULE: when you add the ln, the exponent comes down in front as so.  *now take the derivative of both sides! 1 dy = x(0)+ln2(1) y dx  Dy/dx=ln2(y)*plug in your original equation for y  Dy/dx=ln2(2^x)

47. Formula: *y = a^x Dy/dx = a^x lna (this would help with the first bullet on the previous slide)

48. Now your turn  F(x)=x^sinx

49. F(x)=x^sinx  Lny=sinxlnx  1 dy = sinx 1 (1) + lnxcosx x dx x  Dy/dx= x^sinx[sinx/x + lnxcosx] ^we mult. By y!

50. Almost done... Now let’s try an FRQ  1975 AB2  This FRQ is about particle motion, which involves position, velocity and acceleration.  In order to solve these kind of problems you must be able to take the derivative!  For the derivative of position=velocity  And the derivative of velocity=acceleration

51. 1975 AB 2  Given the function f defined by f(x)=ln(x 2 -9).  A. Describe the symmetry of the graph of f.  B. Find the domain of f.  C. Find all values of x such that f(x)=O.  D. Write a formula for f -1 (x), the inverse function of f, for x >3.

52. © Laura Berglind and Avery Gibson, 2011, AP Calculus AB, Autrey THE END