implicit differentiation, part 2
DESCRIPTION
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.htmlTRANSCRIPT
![Page 1: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/1.jpg)
![Page 2: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/2.jpg)
![Page 3: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/3.jpg)
Example 2
![Page 4: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/4.jpg)
Example 2
Let’s find y ′ given that:
![Page 5: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/5.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
![Page 6: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/6.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
![Page 7: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/7.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
![Page 8: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/8.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)− 5
dy
dx=
d
dx
(x3)
![Page 9: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/9.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5dy
dx=
d
dx
(x3)
![Page 10: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/10.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=
d
dx
(x3)
![Page 11: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/11.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
![Page 12: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/12.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
![Page 13: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/13.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
4.
![Page 14: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/14.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
4.(2x).
![Page 15: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/15.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
4.(2x).y+
![Page 16: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/16.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
4.(2x).y + 4x2.
![Page 17: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/17.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
4.(2x).y + 4x2.y ′−
![Page 18: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/18.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
4.(2x).y + 4x2.y ′ − 5y ′ =
![Page 19: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/19.jpg)
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
d
dx
[4x2y − 5y
]=
d
dx
(x3)
d
dx
(4x2y
)︸ ︷︷ ︸product rule!
−5����y ′
dy
dx=�����>
3x2
d
dx
(x3)
Here we apply the product rule:
4.(2x).y + 4x2.y ′ − 5y ′ = 3x2
![Page 20: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/20.jpg)
Example 2
![Page 21: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/21.jpg)
Example 2
We have the relation:
![Page 22: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/22.jpg)
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
![Page 23: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/23.jpg)
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
We only need to solve for y ′:
![Page 24: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/24.jpg)
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
We only need to solve for y ′:
y ′(4x2 − 5
)= 3x2 − 8xy
![Page 25: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/25.jpg)
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
We only need to solve for y ′:
y ′(4x2 − 5
)= 3x2 − 8xy
y ′ =3x2 − 8xy
4x2 − 5
![Page 26: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/26.jpg)
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
We only need to solve for y ′:
y ′(4x2 − 5
)= 3x2 − 8xy
y ′ =3x2 − 8xy
4x2 − 5
![Page 27: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/27.jpg)
Example 3
![Page 28: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/28.jpg)
Example 3
Let’s consider the equation:
![Page 29: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/29.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
![Page 30: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/30.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
![Page 31: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/31.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=
d
dx
(a
23
)
![Page 32: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/32.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)
![Page 33: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/33.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
![Page 34: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/34.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
![Page 35: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/35.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
2
3.
![Page 36: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/36.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
2
3.x
23−1+
![Page 37: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/37.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
2
3.x
23−1 +
2
3.
![Page 38: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/38.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
2
3.x
23−1 +
2
3.y
23−1.
![Page 39: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/39.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
2
3.x
23−1 +
2
3.y
23−1.y ′ = 0
![Page 40: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/40.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
2
3.x
23−1 +
2
3.y
23−1.y ′ = 0
2
3.x−
13 +
2
3.y−
13 .y ′ = 0
![Page 41: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/41.jpg)
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
d
dx
[x
23 + y
23
]=��
���*
0d
dx
(a
23
)d
dx
(x
23
)+
d
dx
(y
23
)= 0
Here we apply the chain rule:
2
3.x
23−1 +
2
3.y
23−1.y ′ = 0
���2
3.x−
13 +���2
3.y−
13 .y ′ = 0
![Page 42: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/42.jpg)
Example 3
![Page 43: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/43.jpg)
Example 3
We now have the equation:
![Page 44: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/44.jpg)
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
![Page 45: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/45.jpg)
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
We just solve for y ′:
![Page 46: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/46.jpg)
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
We just solve for y ′:
y−13 y ′ = −x−
13
![Page 47: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/47.jpg)
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
We just solve for y ′:
y−13 y ′ = −x−
13
y ′ = −x−13
y−13
![Page 48: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/48.jpg)
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
We just solve for y ′:
y−13 y ′ = −x−
13
y ′ = −x−13
y−13
![Page 49: Implicit Differentiation, Part 2](https://reader034.vdocuments.us/reader034/viewer/2022042521/557e78a0d8b42a03668b512a/html5/thumbnails/49.jpg)