In this presentation we solve two more examples of implicit differentiation problems. We use a faster, more direct method.
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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
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