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
Example 2
Example 2
Let’s find y ′ given that:
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
Example 2
Let’s find y ′ given that:
4x2y − 5y = x3
First, we apply the derivative to both sides:
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)
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)
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)
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)
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)
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:
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.
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).
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+
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.
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 ′−
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 ′ =
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
Example 2
Example 2
We have the relation:
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
We only need to solve for y ′:
Example 2
We have the relation:
8xy + 4x2y ′ − 5y ′ = 3x2
We only need to solve for y ′:
y ′(4x2 − 5
)= 3x2 − 8xy
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
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
Example 3
Example 3
Let’s consider the equation:
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Example 3
Let’s consider the equation:
x23 + y
23 = a
23
Let’s find y ′:
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
)
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
)
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
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:
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.
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+
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.
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.
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
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
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
Example 3
Example 3
We now have the equation:
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
We just solve for y ′:
Example 3
We now have the equation:
x−13 + y−
13 y ′ = 0
We just solve for y ′:
y−13 y ′ = −x−
13
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
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
