The Chain Rule, Part 1

The Chain Rule
The chain rule tells us what is the derivative of a composite
function.

The Chain Rule
function.
For example, let’s consider this function:

The Chain Rule
function.
h(x) = sin x2

The Chain Rule
function.
h(x) = sin x2
This is the composition of two functions:

The Chain Rule
function.
h(x) = sin x2
f (y) = sin y g(x) = x2

The Chain Rule
function.
h(x) = sin x2
f (y) = sin y g(x) = x2
That is:

The Chain Rule
function.
h(x) = sin x2
f (y) = sin y g(x) = x2
That is:
h(x) = f [g(x)]

The Chain Rule
function.
h(x) = sin x2
f (y) = sin y g(x) = x2
That is:
h(x) = f [g(x)]
How do we ﬁnd the derivative of h?

The Chain Rule
function.
h(x) = sin x2
f (y) = sin y g(x) = x2
That is:
h(x) = f [g(x)]
How do we ﬁnd the derivative of h?
The answer is: The Chain Rule!

The Chain Rule
The chain rule tells us that, if we have a composite function:

The Chain Rule
h(x) = f [g(x)]

The Chain Rule
h(x) = f [g(x)]
Its derivative is:

The Chain Rule
h(x) = f [g(x)]
Its derivative is:
h (x) = f [g(x)] g (x)

Example 1
Let’s ﬁnd the derivative of:

Example 1
h(x) = sin x2

Example 1
h(x) = sin x2
This is the composition of the functions:

Example 1
h(x) = sin x2
f (y) = sin y g(x) = x2

Example 1
h(x) = sin x2
f (y) = sin y g(x) = x2
h(x) = f [g(x)]

Example 1
h(x) = sin x2
f (y) = sin y g(x) = x2
h(x) = f [g(x)]
The chain rule says that:

Example 1
h(x) = sin x2
f (y) = sin y g(x) = x2
h(x) = f [g(x)]
h (x) = f [g(x)] g (x)

Example 1
h(x) = sin x2
f (y) = sin y g(x) = x2
h(x) = f [g(x)]
h (x) = f [g(x)] g (x)
And we have that:

Example 1
h(x) = sin x2
f (y) = sin y g(x) = x2
h(x) = f [g(x)]
h (x) = f [g(x)] g (x)
And we have that:
f (y) = cos y

Example 1
h(x) = sin x2
f (y) = sin y g(x) = x2
h(x) = f [g(x)]
h (x) = f [g(x)] g (x)
And we have that:
f (y) = cos y g (x) = 2x

Example 1
Evaluating f at g(x):

Example 1
f [g(x)] = cos g(x) = cos x2

Example 1
So:

Example 1
So:
h (x) =

Example 1
So:
h (x) = cos x2
.

Example 1
So:
h (x) =
f [g(x)]
cos x2
.

Example 1
So:
h (x) = cos x2
.2x

Example 1
So:
h (x) = cos x2
.
g (x)
2x

Example 1
So:
h (x) = cos x2
.2x = 2x cos x2

Example 2
h(x) =
√
1 + x
Here, h is a composition of the functions:

Example 2
h(x) =
√
1 + x
f (y) =
√
y

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:
h (x) = f [g(x)] g (x)

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:
h (x) = f [g(x)] g (x) =
1
2
.g(x)−1/2
.

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:
h (x) = f [g(x)] g (x) =
f [g(x)]
1
2
.g(x)−1/2
.

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:
h (x) = f [g(x)] g (x) =
1
2
.g(x)−1/2
.1

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:
h (x) = f [g(x)] g (x) =
1
2
.g(x)−1/2
.
g (x)
1

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:
h (x) = f [g(x)] g (x) =
1
2
.g(x)−1/2
.1 =
1
2 g(x)

Example 2
h(x) =
√
1 + x
f (y) =
√
y = y
1
2 g(x) = 1 + x
f (y) =
1
2
y−1
2 g (x) = 1
We have that:
h (x) = f [g(x)] g (x) =
1
2
.g(x)−1/2
.1 =
1
2 g(x)
=
1
2
√
1 + x

The Chain Rule, Part 1

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The Chain Rule, Part 1