implicit differentiation, part 1
TRANSCRIPT
The Idea of Implicit Differentiation
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions.
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.What if we have a relation:
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.What if we have a relation:
x2 + y2 = 25
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.What if we have a relation:
x2 + y2 = 25
Solving for y , we get a function of x :
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.What if we have a relation:
x2 + y2 = 25
Solving for y , we get a function of x :
y =√
25− x2
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.What if we have a relation:
x2 + y2 = 25
Solving for y , we get a function of x :
y =√
25− x2
Implicit differentiation allows you to find the derivative of y(x).
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.What if we have a relation:
x2 + y2 = 25
Solving for y , we get a function of x :
y =√
25− x2
Implicit differentiation allows you to find the derivative of y(x).Without the pain of solving the equation for y !
The Idea of Implicit Differentiation
Until now, we only dealt with explicit functions. For example:
f (x) = x2
Given x , you calculate f (x) directly.What if we have a relation:
x2 + y2 = 25
Solving for y , we get a function of x :
y =√
25− x2
Implicit differentiation allows you to find the derivative of y(x).Without the pain of solving the equation for y ! (There are casesyou can’t do that).
Example 1
Example 1
Let’s stick with the previous relation:
Example 1
Let’s stick with the previous relation:
x2 + y2 = 25
Example 1
Let’s stick with the previous relation:
x2 + y2 = 25
First of all, let’s write:
Example 1
Let’s stick with the previous relation:
x2 + y2 = 25
First of all, let’s write:
y = f (x)
Example 1
Let’s stick with the previous relation:
x2 + y2 = 25
First of all, let’s write:
y = f (x)
x2 + [f (x)]2 = 25
Example 1
Example 1
Let’s make the following notational distinction:
Example 1
Let’s make the following notational distinction:
dy
dx
Example 1
Let’s make the following notational distinction:
dy
dx︸︷︷︸noun
Example 1
Let’s make the following notational distinction:
dy
dx︸︷︷︸noun
= The derivative of y
Example 1
Let’s make the following notational distinction:
dy
dx︸︷︷︸noun
= The derivative of y
d
dy
Example 1
Let’s make the following notational distinction:
dy
dx︸︷︷︸noun
= The derivative of y
d
dy︸︷︷︸verb
Example 1
Let’s make the following notational distinction:
dy
dx︸︷︷︸noun
= The derivative of y
d
dx︸︷︷︸verb
= Take the derivative!
Example 1
Example 1
Now, let’s return to our problem:
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
d
dx
[x2 + [f (x)]2
]=
d
dx(25)
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
d
dx
[x2 + [f (x)]2
]=
d
dx(25)
d
dx
(x2)
+d
dx[f (x)]2 =
d
dx(25)
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
d
dx
[x2 + [f (x)]2
]=
d
dx(25)
�����>
2xd
dx
(x2)
+d
dx[f (x)]2 =
d
dx(25)
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
d
dx
[x2 + [f (x)]2
]=
d
dx(25)
�����>
2xd
dx
(x2)
+d
dx[f (x)]2 =
����>
0d
dx(25)
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
d
dx
[x2 + [f (x)]2
]=
d
dx(25)
�����>
2xd
dx
(x2)
+d
dx[f (x)]2 =
����>
0d
dx(25)
2x +d
dx[f (x)]2 = 0
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
d
dx
[x2 + [f (x)]2
]=
d
dx(25)
�����>
2xd
dx
(x2)
+d
dx[f (x)]2 =
����>
0d
dx(25)
2x +d
dx[f (x)]2︸ ︷︷ ︸
How do we solve this?
= 0
Example 1
Now, let’s return to our problem:
x2 + [f (x)]2 = 25
This is an equality between two functions.So, their derivatives must be equal:
d
dx
[x2 + [f (x)]2
]=
d
dx(25)
�����>
2xd
dx
(x2)
+d
dx[f (x)]2 =
����>
0d
dx(25)
2x +d
dx[f (x)]2︸ ︷︷ ︸
How do we solve this?
= 0
The chain rule!
Example 1
Example 1
The chain rule is the secret of implicit differentiation.
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 =
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).f ′(x)
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).f ′(x)
So:
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).f ′(x)
So:
2x + 2f (x)f ′(x) = 0
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).f ′(x)
So:
2x + 2f (x)f ′(x) = 0
And now we can solve for f ′(x):
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).f ′(x)
So:
2x + 2f (x)f ′(x) = 0
And now we can solve for f ′(x):
2f (x)f ′(x) = −2x
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).f ′(x)
So:
2x + 2f (x)f ′(x) = 0
And now we can solve for f ′(x):
�2f (x)f ′(x) = −�2x
Example 1
The chain rule is the secret of implicit differentiation.
2x +d
dx[f (x)]2 = 0
This derivative we can solve applying the chain rule:
d
dx[f (x)]2 = 2.f (x).f ′(x)
So:
2x + 2f (x)f ′(x) = 0
And now we can solve for f ′(x):
�2f (x)f ′(x) = −�2x
f ′(x) = − x
f (x)
Example 1
Example 1
But remember that f (x) = y :
Example 1
But remember that f (x) = y :
y ′ = −x
y
Example 1
But remember that f (x) = y :
y ′ = −x
y
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
y =√
25− x2
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
y =√
25− x2
We can apply the chain rule:
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
y =√
25− x2
We can apply the chain rule:
y ′ =1
2.
1√25− x2
. (−2x)
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
y =√
25− x2
We can apply the chain rule:
y ′ =1
�2.
1√25− x2
. (−�2x)
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
y =√
25− x2
We can apply the chain rule:
y ′ =1
�2.
1
������: y√
25− x2. (−�2x)
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
y =√
25− x2
We can apply the chain rule:
y ′ =1
�2.
1
������: y√
25− x2. (−�2x)
y ′ = −x
y
Example 1
But remember that f (x) = y :
y ′ = −x
y
Let’s try to verify that:
x2 + y2 = 25
y =√
25− x2
We can apply the chain rule:
y ′ =1
�2.
1
������: y√
25− x2. (−�2x)
y ′ = −x
y
