LinkedIn emplea cookies para mejorar la funcionalidad y el rendimiento de nuestro sitio web, así como para ofrecer publicidad relevante. Si continúas navegando por ese sitio web, aceptas el uso de cookies. Consulta nuestras Condiciones de uso y nuestra Política de privacidad para más información.

LinkedIn emplea cookies para mejorar la funcionalidad y el rendimiento de nuestro sitio web, así como para ofrecer publicidad relevante. Si continúas navegando por ese sitio web, aceptas el uso de cookies. Consulta nuestra Política de privacidad y nuestras Condiciones de uso para más información.

Se ha denunciado esta presentación.

¿Recomiendas esta presentación? ¿Por qué no compartes?

- Summary of exposure draft on Hedge ... by Venkata Subramani... 2832 views
- Kalkulus II (23 - 24) by jayamartha 1002 views
- Literature Review - Breadfruit by Nawsheen Hosenally 10701 views
- Derivatives market by Nikhiliit 83628 views
- Financial derivatives ppt by Logasakthi Kandasamy 63514 views
- Financial derivatives ppt by VaishnaviSavant 146789 views

538 visualizaciones

Publicado el

Sin descargas

Visualizaciones totales

538

En SlideShare

0

De insertados

0

Número de insertados

2

Compartido

0

Descargas

0

Comentarios

0

Recomendaciones

1

No insertados

No hay notas en la diapositiva.

- 1. Review on Derivatives http://www.lahc.edu/math/frankma.htm
- 2. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined.
- 3. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined. P y = f(x)
- 4. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined. This implies that the graph is smooth at P. P y = f(x)
- 5. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined. This implies that the graph is smooth at P. A function such as y = | x | is not differentiable at (0, 0) because of the abrupt change at that point. P y = f(x)
- 6. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined. This implies that the graph is smooth at P. A function such as y = | x | is not differentiable at (0, 0) because of the abrupt change at that point. P y = f(x) y = |x|
- 7. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined. This implies that the graph is smooth at P. A function such as y = | x | is not differentiable at (0, 0) because of the abrupt change at that point. A function is differentiable over an interval (a, b) if it’s differentiable at all points in the interval. P y = f(x) y = |x|
- 8. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined. This implies that the graph is smooth at P. A function such as y = | x | is not differentiable at (0, 0) because of the abrupt change at that point. y = f(x) P y = |x| A function is differentiable over an interval (a, b) if it’s differentiable at all points in the interval. P a y = f(x) b
- 9. Review on Derivatives A function f(x) is differentiable at a point P on its graph y = f(x) if the slope (of the tangent) at P is well defined. This implies that the graph is smooth at P. A function such as y = | x | is not differentiable at (0, 0) because of the abrupt change at that point. A function is differentiable over an interval (a, b) if it’s differentiable at all points in the interval. Geometrically, this implies that the graph of y = f(x) a is a smooth curve between a and b. y = f(x) P y = |x| P y = f(x) b
- 10. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x).
- 11. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx
- 12. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx P x y = f(x)
- 13. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx f ’(x) = slope of the tangent at x P x y = f(x)
- 14. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx Rules for Finding the Derivatives f ’(x) = slope of the tangent at x P x y = f(x)
- 15. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx Rules for Finding the Derivatives The slope at a generic point f ’(x) = slope of P = (x, y) is the limit as h 0 the tangent at x of the difference quotient (slope), P x y = f(x)
- 16. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx Rules for Finding the Derivatives The slope at a generic point f ’(x) = slope of P = (x, y) is the limit as h 0 the tangent at x of the difference quotient (slope), P i.e. f ’(x) ≡ lim Δy y = f(x) h 0 Δx x
- 17. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx Rules for Finding the Derivatives The slope at a generic point f ’(x) = slope of P = (x, y) is the limit as h 0 the tangent at x of the difference quotient (slope), P i.e. f ’(x) ≡ lim Δy y = f(x) h 0 Δx ≡ lim f(x + h) – f(x) x h0 h
- 18. Review of Derivatives The slopes–formula of the graph for y = f(x) may be “derived” from f(x). Therefore it’s called the derivative df . of f(x) and it’s denoted as f ’(x) or dx Rules for Finding the Derivatives The slope at a generic point f ’(x) = slope of P = (x, y) is the limit as h 0 the tangent at x of the difference quotient (slope), P i.e. f ’(x) ≡ lim Δy y = f(x) h 0 Δx ≡ lim f(x + h) – f(x) x h0 h The existence of this limit means that we are able to “balance” a tangent line at P and that the limit is the slope of this line.
- 19. Review of Derivatives Elementary mathematics functions are functions built from three basic families.
- 20. Review of Derivatives Elementary mathematics functions are functions built from three basic families. * the power functions xp
- 21. Review of Derivatives Elementary mathematics functions are functions built from three basic families. * the power functions xp * the trig. functions such as sin(x), cos(x) and their inverses arcsin(x), arccos(x)
- 22. Review of Derivatives Elementary mathematics functions are functions built from three basic families. * the power functions xp * the trig. functions such as sin(x), cos(x) and their inverses arcsin(x), arccos(x) * the log–function ln(x) and exponential functions ex.
- 23. Review of Derivatives Elementary mathematics functions are functions built from three basic families. * the power functions xp * the trig. functions such as sin(x), cos(x) and their inverses arcsin(x), arccos(x) * the log–function ln(x) and exponential functions ex. New formulas are constructed from these basic ones, with real numbers, using the algebraic operations +, – , *, / , and the composition operation “○” of functions.
- 24. Review of Derivatives Elementary mathematics functions are functions built from three basic families. * the power functions xp * the trig. functions such as sin(x), cos(x) and their inverses arcsin(x), arccos(x) * the log–function ln(x) and exponential functions ex. New formulas are constructed from these basic ones, with real numbers, using the algebraic operations +, – , *, / , and the composition operation “○” of functions. Hence, to calculate derivatives systematically we find I. the derivatives of the basic functions,
- 25. Review of Derivatives Elementary mathematics functions are functions built from three basic families. * the power functions xp * the trig. functions such as sin(x), cos(x) and their inverses arcsin(x), arccos(x) * the log–function ln(x) and exponential functions ex. New formulas are constructed from these basic ones, with real numbers, using the algebraic operations +, – , *, / , and the composition operation “○” of functions Hence, to calculate derivatives systematically we find I. the derivatives of the basic functions, II. understand how the operations +, – , *, / and function composition behave under differentiation.
- 26. Review of Derivatives The basic derivatives that we know up to now are
- 27. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp)
- 28. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) (Derivatives of the trigonometric functions) dsin(x) dcos(x) dx dx dtan(x) dcot(x) dx dx dsec(x) dcsc(x) dx dx
- 29. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) dx dx dtan(x) dcot(x) dx dx dsec(x) dcsc(x) dx dx
- 30. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) dx = cos(x) dx dtan(x) dcot(x) dx dx dsec(x) dcsc(x) dx dx
- 31. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) dx dx dsec(x) dcsc(x) dx dx
- 32. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) 2(x) dx = sec dx dsec(x) dcsc(x) dx dx
- 33. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) 2(x) = –csc2(x) dx = sec dx dsec(x) dcsc(x) dx dx
- 34. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) 2(x) = –csc2(x) dx = sec dx dsec(x) dcsc(x) dx = sec(x) tan(x) dx
- 35. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) 2(x) = –csc2(x) dx = sec dx dsec(x) dcsc(x) = –csc(x)cot(x) dx = sec(x) tan(x) dx
- 36. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) 2(x) = –csc2(x) dx = sec dx dsec(x) dcsc(x) = –csc(x)cot(x) dx = sec(x) tan(x) dx Derivatives “respect” the operations of +, – , and constant multiplication.
- 37. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) 2(x) = –csc2(x) dx = sec dx dsec(x) dcsc(x) = –csc(x)cot(x) dx = sec(x) tan(x) dx Derivatives “respect” the operations of +, – , and constant multiplication. But for the operations *, /, and function–composition, the derivative behaves differently.
- 38. Review of Derivatives The basic derivatives that we know up to now are (Derivatives of xp) p p then f ’(x) = pxp–1 or dx = pxp–1 If f(x) = x dx (Derivatives of the trigonometric functions) dsin(x) dcos(x) = –sin(x) dx = cos(x) dx dtan(x) dcot(x) 2(x) = –csc2(x) dx = sec dx dsec(x) dcsc(x) = –csc(x)cot(x) dx = sec(x) tan(x) dx Derivatives “respect” the operations of +, – , and constant multiplication. But for the operations *, /, and function–composition, the derivative behaves differently. Specifically, derivative of composed–function obeys the Chain Rule.
- 39. Review of Derivatives (Derivative Rules for ±, and Constant Multiplication.) Let f = f(x), g = g(x) and c be a constant and that the prime ( )’ operation means derivative taken with respect to x. a. (cf) ’ = c(f) ’. b. (f ± g) ’ = (f) ’ ± (g) ’.
- 40. Review of Derivatives (Derivative Rules for ±, and Constant Multiplication.) Let f = f(x), g = g(x) and c be a constant and that the prime ( )’ operation means derivative taken with respect to x. a. (cf) ’ = c(f) ’. b. (f ± g) ’ = (f) ’ ± (g) ’. These two facts may be phrased together as (af ± bg) ’ = a(f) ’ ± b(g)’ where a and b are numbers.
- 41. Review of Derivatives (Derivative Rules for ±, and Constant Multiplication.) Let f = f(x), g = g(x) and c be a constant and that the prime ( )’ operation means derivative taken with respect to x. a. (cf) ’ = c(f) ’. b. (f ± g) ’ = (f) ’ ± (g) ’. These two facts may be phrased together as (af ± bg) ’ = a(f) ’ ± b(g)’ where a and b are numbers. Example A. (2sin(x) + x – 2) a. Find the derivative of 3
- 42. Review of Derivatives (Derivative Rules for ±, and Constant Multiplication.) Let f = f(x), g = g(x) and c be a constant and that the prime ( )’ operation means derivative taken with respect to x. a. (cf) ’ = c(f) ’. b. (f ± g) ’ = (f) ’ ± (g) ’. These two facts may be phrased together as (af ± bg) ’ = a(f) ’ ± b(g)’ where a and b are numbers. Example A. (2sin(x) + x – 2) a. Find the derivative of 3 (2sin(x) + x – 2) ]’ [ 3
- 43. Review of Derivatives (Derivative Rules for ±, and Constant Multiplication.) Let f = f(x), g = g(x) and c be a constant and that the prime ( )’ operation means derivative taken with respect to x. a. (cf) ’ = c(f) ’. b. (f ± g) ’ = (f) ’ ± (g) ’. These two facts may be phrased together as (af ± bg) ’ = a(f) ’ ± b(g)’ where a and b are numbers. Example A. (2sin(x) + x – 2) a. Find the derivative of 3 (2sin(x) + x – 2) 1 [2sin(x) + x – 2] ’ ]’ = 3 [ 3
- 44. Review of Derivatives (Derivative Rules for ±, and Constant Multiplication.) Let f = f(x), g = g(x) and c be a constant and that the prime ( )’ operation means derivative taken with respect to x. a. (cf) ’ = c(f) ’. b. (f ± g) ’ = (f) ’ ± (g) ’. These two facts may be phrased together as (af ± bg) ’ = a(f) ’ ± b(g)’ where a and b are numbers. Example A. (2sin(x) + x – 2) a. Find the derivative of 3 (2sin(x) + x – 2) 1 [2sin(x) + x – 2] ’ ]’ = 3 [ 3 = 1 (2cos(x) + 1) 3
- 45. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3
- 46. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3).
- 47. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0)
- 48. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 3
- 49. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 = 1 3 So by the tangent line formula y = f(x0)’(x – x0) + y0 we have the tangent line as y = x – 2/3.
- 50. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 = 1 3 So by the tangent line formula y = f(x0)’(x – x0) + y0 we have the tangent line as y = x – 2/3. Graphically, there are 4 possibilities at P = (0, –2/3).
- 51. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 = 1 3 So by the tangent line formula y = f(x0)’(x – x0) + y0 we have the tangent line as y = x – 2/3. Graphically, there are 4 possibilities at P = (0, –2/3). P
- 52. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 = 1 3 So by the tangent line formula y = f(x0)’(x – x0) + y0 we have the tangent line as y = x – 2/3. Graphically, there are 4 possibilities at P = (0, –2/3). P P
- 53. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 = 1 3 So by the tangent line formula y = f(x0)’(x – x0) + y0 we have the tangent line as y = x – 2/3. Graphically, there are 4 possibilities at P = (0, –2/3). P P P
- 54. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 = 1 3 So by the tangent line formula y = f(x0)’(x – x0) + y0 we have the tangent line as y = x – 2/3. Graphically, there are 4 possibilities at P = (0, –2/3). P P P P
- 55. Review of Derivatives b. Find the slope and the equation of the tangent line (2sin(x) + x – 2) at x = 0 for f(x) = . 3 At x = 0, f(0) = –2/3, hence the tangent line is based at (0, –2/3). The slope of the tangent is f ’(0) = 1 (2cos(x) + 1) | x=0 = 1 3 So by the tangent line formula y = f(x0)’(x – x0) + y0 we have the tangent line as y = x – 2/3. Graphically, there are 4 possibilities at P = (0, –2/3). P P P P Ex. Which is it (without graphing it or using the 2nd derivative)?
- 56. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = f (g )’=
- 57. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) f (g )’=
- 58. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) f (g = g(f)’ – f(g)’ (The Quotient Rule) )’ g2
- 59. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. a. x2cos(x) f (g
- 60. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. a. x2cos(x) [x2cos(x)]’ f (g
- 61. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. a. x2cos(x) [x2cos(x)]’ = (x2)’cos(x) + x2[cos(x)]’ f (g
- 62. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. a. x2cos(x) [x2cos(x)]’ = (x2)’cos(x) + x2[cos(x)]’ = 2xcos(x) – x2sin(x) f (g
- 63. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. 2cos(x) x2 a. x b. cos(x) [x2cos(x)]’ = (x2)’cos(x) + x2[cos(x)]’ = 2xcos(x) – x2sin(x) f (g
- 64. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. 2cos(x) x2 a. x b. cos(x) [x2cos(x)]’ x2 ]’ = (x2)’cos(x) + x2[cos(x)]’ [ cos(x) = 2xcos(x) – x2sin(x) f (g
- 65. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. 2cos(x) x2 a. x b. cos(x) [x2cos(x)]’ x2 ]’ = (x2)’cos(x) + x2[cos(x)]’ [ cos(x) = 2xcos(x) – x2sin(x) = cos2(x) f (g
- 66. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. 2cos(x) x2 a. x b. cos(x) [x2cos(x)]’ x2 ]’ = (x2)’cos(x) + x2[cos(x)]’ [ cos(x) = 2xcos(x) – x2sin(x) cos(x)(x2)’ = cos2(x) f (g
- 67. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. 2cos(x) x2 a. x b. cos(x) [x2cos(x)]’ x2 ]’ = (x2)’cos(x) + x2[cos(x)]’ [ cos(x) = 2xcos(x) – x2sin(x) cos(x)(x2)’ – (cos(x))’x2 = cos2(x) f (g
- 68. Review of Derivatives (The Product and Quotient Rules for Derivative .) Let f = f(x), g = g(x) and that the prime ( )’ operation means derivative taken with respect to x, then (fg)’ = (f)’g + f(g)’ (The Product Rule) = g(f)’ – f(g)’ (The Quotient Rule) )’ g2 Example B. Find the derivative. 2cos(x) x2 a. x b. cos(x) [x2cos(x)]’ x2 ]’ = (x2)’cos(x) + x2[cos(x)]’ [ cos(x) = 2xcos(x) – x2sin(x) cos(x)(x2)’ – (cos(x))’x2 = cos2(x) 2xcos(x) + x2sin(x) = cos2(x) f (g
- 69. Review of Derivatives (Derivative Rule for Composition–The Chain Rule)
- 70. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition.
- 71. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u.
- 72. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5,
- 73. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3.
- 74. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3. The composition g(f(x)) where y is a function of x is y = 3(5x) = 15x
- 75. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3. The composition g(f(x)) where y is a function of x is y = 3(5x) = 15x which has a slope of (3)(5) = 15.
- 76. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3. The composition g(f(x)) where y is a function of x is y = 3(5x) = 15x which has a slope of (3)(5) = 15. (The Chain Rule) Let u = f(x) and y = g(u) so that y = g(f(x)), then dy = dy du dx du dx
- 77. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3. The composition g(f(x)) where y is a function of x is y = 3(5x) = 15x which has a slope of (3)(5) = 15. (The Chain Rule) Let u = f(x) and y = g(u) so that y = g(f(x)), then dy = dy du dx du dx or [g(f(x))]’ = g’(f(x)) * f ’(x)
- 78. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3. The composition g(f(x)) where y is a function of x is y = 3(5x) = 15x which has a slope of (3)(5) = 15. (The Chain Rule) Let u = f(x) and y = g(u) so that y = g(f(x)), then dy = dy du dx du dx derivative taken with respect to x or [g(f(x))]’ = g’(f(x)) * f ’(x)
- 79. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3. The composition g(f(x)) where y is a function of x is y = 3(5x) = 15x which has a slope of (3)(5) = 15. (The Chain Rule) Let u = f(x) and y = g(u) so that y = g(f(x)), then dy = dy du dx du dx derivative taken with respect to x or derivative taken with respect to u [g(f(x))]’ = g’(f(x)) * f ’(x)
- 80. Review of Derivatives (Derivative Rule for Composition–The Chain Rule) The Chain Rule gives the multiplying effect of the rates of change under composition. For instance, let u = f(x) = 5x and y = g(u) = 3u. The slope of u = 5x is 5, and the slope of y = g(u) = 3u as a function of u is 3. The composition g(f(x)) where y is a function of x is y = 3(5x) = 15x which has a slope of (3)(5) = 15. (The Chain Rule) Let u = f(x) and y = g(u) so that y = g(f(x)), then dy = dy du dx du dx derivative taken with respect to x or derivative taken with respect to u derivative taken with respect to x [g(f(x))]’ = g’(f(x)) * f ’(x)
- 81. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx.
- 82. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2),
- 83. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx
- 84. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx
- 85. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx = 2xcos(u)
- 86. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx = 2xcos(u) = 2xcos(x2)
- 87. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx = 2xcos(u) = 2xcos(x2) Applying the chain rule to the basic functions yields useful specific rules for calculation.
- 88. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx = 2xcos(u) = 2xcos(x2) Applying the chain rule to the basic functions yields useful specific rules for calculation. For example, let y = sin(u) and u = u(x) so that y = sin(u(x)),
- 89. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx = 2xcos(u) = 2xcos(x2) Applying the chain rule to the basic functions yields useful specific rules for calculation. For example, let y = sin(u) and u = u(x) so that y = sin(u(x)), then [sin(u(x))]’ derivative taken with respect to x
- 90. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx = 2xcos(u) = 2xcos(x2) Applying the chain rule to the basic functions yields useful specific rules for calculation. For example, let y = sin(u) and u = u(x) so that y = sin(u(x)), then [sin(u(x))]’ = [sin(u)]’ * u(x)’ derivative taken derivative taken with respect to x with respect to u derivative taken with respect to u
- 91. Review of Derivatives Example C. Let u = x2 and y = sin(u), express y as a function of x and find dy/dx. y = sin(u) = sin(x2), dy dy du dx = du dx dsin(u) dx2 = du dx = 2xcos(u) = 2xcos(x2) Applying the chain rule to the basic functions yields useful specific rules for calculation. For example, let y = sin(u) and u = u(x) so that y = sin(u(x)), then [sin(u(x))]’ = [sin(u)]’ * u(x)’ = cos(u)*(u)’. derivative taken derivative taken with respect to x with respect to u derivative taken with respect to u
- 92. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas .
- 93. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = b. [cos(u)]’ = c. [up] =
- 94. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = c. [up] =
- 95. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] =
- 96. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’
- 97. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x.
- 98. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x. Example D. Find the derivative of cos[(x2+1)3] (cos[(x2+1)3])’
- 99. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x. Example D. Find the derivative of cos[(x2+1)3] (cos[(x2+1)3])’ u(x)
- 100. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x. Example D. Find the derivative of cos[(x2+1)3] (cos[(x2+1)3])’ = –sin[u] [(x2+1)3] ’ u(x) u’
- 101. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x. Example D. Find the derivative of cos[(x2+1)3] (cos[(x2+1)3])’ = –sin[(x2+1)3] [(x2+1)3]’
- 102. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x. Example D. Find the derivative of cos[(x2+1)3] (cos[(x2+1)3])’ = –sin[(x2+1)3] [(x2+1)3]’ the new u(x) for u3
- 103. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x. Example D. Find the derivative of cos[(x2+1)3] (cos[(x2+1)3])’ = –sin[(x2+1)3] [(x2+1)3]’ the new u(x) for u3 = –sin[(x2+1)3] [3(x2+1)2] (x2+1)’ the new u’
- 104. Review of Derivatives Similarly if u = u(x) and y = cos(u) or up, we get the following specific chain rules formulas . (Specific Chain Rule Formulas) a. [sin(u)]’ = cos(u)*(u)’ b. [cos(u)]’ = –sin(u)*(u)’ c. [up] = pup–1(u)’ where u = u(x) and ( )’ is the derivative with respect to x. Example D. Find the derivative of cos[(x2+1)3] (cos[(x2+1)3])’ = –sin[(x2+1)3] [(x2+1)3]’ the new u(x) for u3 = –sin[(x2+1)3] [3(x2+1)2] (x2+1)’ the new u’ = –sin[(x2+1)3] [3(x2+1)2] 2x

No se han encontrado tableros de recortes públicos para esta diapositiva.

Parece que ya has recortado esta diapositiva en .

Crear un tablero de recortes

Sé el primero en comentar