In this presentation we solve two more examples of implicit differentiation problems. We use a faster, more direct method. For more lessons visit: http://www.intuitive-calculus.com/implicit-differentiation.html

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