Implicit Differentiation, Part 2
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The document presents examples of finding the variable y using derivatives and applying the product rule. It discusses the process step-by-step with mathematical equations, ultimately leading to the calculation of y in different scenarios. The content is instructional, focusing on derivative applications in relation to specific equations.
Implicit Differentiation, Part 2
- 3. Example 2
- 4. Example 2 Let’s find y given that:
- 5. Example 2 Let’s find y given that: 4x2 y − 5y = x3
- 6. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides:
- 7. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3
- 8. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y − 5 dy dx = d dx x3
- 9. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 dy dx = d dx x3
- 10. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = d dx x3
- 11. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3
- 12. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule:
- 13. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule: 4.
- 14. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule: 4.(2x).
- 15. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule: 4.(2x).y+
- 16. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule: 4.(2x).y + 4x2 .
- 17. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule: 4.(2x).y + 4x2 .y −
- 18. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule: 4.(2x).y + 4x2 .y − 5y =
- 19. Example 2 Let’s find y given that: 4x2 y − 5y = x3 First, we apply the derivative to both sides: d dx 4x2 y − 5y = d dx x3 d dx 4x2 y product rule! −5 ¡ ¡ ¡! y dy dx = & & & &&b 3x2 d dx x3 Here we apply the product rule: 4.(2x).y + 4x2 .y − 5y = 3x2
- 20. Example 2
- 21. Example 2 We have the relation:
- 22. Example 2 We have the relation: 8xy + 4x2 y − 5y = 3x2
- 23. Example 2 We have the relation: 8xy + 4x2 y − 5y = 3x2 We only need to solve for y :
- 24. Example 2 We have the relation: 8xy + 4x2 y − 5y = 3x2 We only need to solve for y : y 4x2 − 5 = 3x2 − 8xy
- 25. Example 2 We have the relation: 8xy + 4x2 y − 5y = 3x2 We only need to solve for y : y 4x2 − 5 = 3x2 − 8xy y = 3x2 − 8xy 4x2 − 5
- 26. Example 2 We have the relation: 8xy + 4x2 y − 5y = 3x2 We only need to solve for y : y 4x2 − 5 = 3x2 − 8xy y = 3x2 − 8xy 4x2 − 5
- 27. Example 3
- 28. Example 3 Let’s consider the equation:
- 29. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3
- 30. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y :
- 31. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = d dx a 2 3
- 32. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3
- 33. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0
- 34. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule:
- 35. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule: 2 3 .
- 36. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule: 2 3 .x 2 3 −1 +
- 37. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule: 2 3 .x 2 3 −1 + 2 3 .
- 38. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule: 2 3 .x 2 3 −1 + 2 3 .y 2 3 −1 .
- 39. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule: 2 3 .x 2 3 −1 + 2 3 .y 2 3 −1 .y = 0
- 40. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule: 2 3 .x 2 3 −1 + 2 3 .y 2 3 −1 .y = 0 2 3 .x−1 3 + 2 3 .y−1 3 .y = 0
- 41. Example 3 Let’s consider the equation: x 2 3 + y 2 3 = a 2 3 Let’s find y : d dx x 2 3 + y 2 3 = ¨¨¨ ¨¨B0 d dx a 2 3 d dx x 2 3 + d dx y 2 3 = 0 Here we apply the chain rule: 2 3 .x 2 3 −1 + 2 3 .y 2 3 −1 .y = 0 £ ££2 3 .x−1 3 + £ ££2 3 .y−1 3 .y = 0
- 42. Example 3
- 43. Example 3 We now have the equation:
- 44. Example 3 We now have the equation: x−1 3 + y−1 3 y = 0
- 45. Example 3 We now have the equation: x−1 3 + y−1 3 y = 0 We just solve for y :
- 46. Example 3 We now have the equation: x−1 3 + y−1 3 y = 0 We just solve for y : y−1 3 y = −x−1 3
- 47. Example 3 We now have the equation: x−1 3 + y−1 3 y = 0 We just solve for y : y−1 3 y = −x−1 3 y = − x−1 3 y−1 3
- 48. Example 3 We now have the equation: x−1 3 + y−1 3 y = 0 We just solve for y : y−1 3 y = −x−1 3 y = − x−1 3 y−1 3