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Jan 09, 2023 541 likes •1.01k views

The Chain Rule Given a composite function: The Chain Rule Given a composite function: h ( x ) = f [ g ( x )] The Chain Rule Given a composite function: h ( x ) = f [ g ( x )] The chain rule says that: The Chain Rule Given a composite

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The Chain Rule

Given a composite function:

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The Chain Rule

Given a composite function: h(x) = f [g(x)]

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that:

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x)

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again:

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin

uter function

x2

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2

inner function

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2 h′(x) = cos x2.

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2 h′(x) = cos x2

derivative of outer function

.

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2 h′(x) = cos x2.

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2 h′(x) = cos x2.2x

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2 h′(x) = cos x2. 2x

derivative of inner function

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The Chain Rule

Given a composite function: h(x) = f [g(x)] The chain rule says that: h′(x) = f ′ [g(x)] g′(x) Let’s consider again: h(x) = sin x2 h′(x) = cos x2.2x

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Example 1

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Example 1

f (x) = 3 √ sin x

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Example 1

f (x) = 3 √ sin x We can write it as:

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3 (sin x)

1 2

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3   sin x

inner function

 

1 2

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

We take the derivative of the outer function:

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

We take the derivative of the outer function: f ′(x) = 3.1 2y− 1

2 .

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

We take the derivative of the outer function: f ′(x) = 3.1 2y− 1

2 .y′

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

We take the derivative of the outer function: f ′(x) = 3.1 2y− 1

2 .y′ = 3

2 (sin x)− 1

2 . (sin x)′

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

We take the derivative of the outer function: f ′(x) = 3.1 2y− 1

2 .y′ = 3

2 (sin x)− 1

2 . (sin x)′ = 3

2 (sin x)− 1

2 . cos x

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

We take the derivative of the outer function: f ′(x) = 3.1 2y− 1

2 .y′ = 3

2 (sin x)− 1

2 . (sin x)′ = 3

2 (sin x)− 1

2 . cos x

f ′(x) = 3 cos x √ sin x

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Example 1

f (x) = 3 √ sin x We can write it as: f (x) = 3

✘✘

✘ ✿y

sin x 1

2 = 3y 1 2

We take the derivative of the outer function: f ′(x) = 3.1 2y− 1

2 .y′ = 3

2 (sin x)− 1

2 . (sin x)′ = 3

2 (sin x)− 1

2 . cos x

f ′(x) = 3 cos x √ sin x

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Example 2

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Example 2

f (x) = a √ sin 2x

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

f ′(x) = a.1

2. (sin 2x)− 1

2 .

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

f ′(x) = a.1

2. (sin 2x)− 1

2 . (sin 2x)′

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

f ′(x) = a.1

2. (sin 2x)− 1

2 . (sin 2x)′

f ′(x) = a.1

2. (sin 2x)− 1

2 . cos 2x.

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

f ′(x) = a.1

2. (sin 2x)− 1

2 . (sin 2x)′

f ′(x) = a.1

2. (sin 2x)− 1

2 . cos 2x.2

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

f ′(x) = a.1

2. (sin 2x)− 1

2 . (sin 2x)′

f ′(x) = a.1

✁

2. (sin 2x)− 1

2 . cos 2x.✁

2

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

f ′(x) = a.1

2. (sin 2x)− 1

2 . (sin 2x)′

f ′(x) = a.1

✁

2. (sin 2x)− 1

2 . cos 2x.✁

2 f ′(x) = a cos 2x √ sin 2x

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Example 2

f (x) = a √ sin 2x = a (sin 2x)

1 2

f ′(x) = a.1

2. (sin 2x)− 1

2 . (sin 2x)′

f ′(x) = a.1

✁

2. (sin 2x)− 1

2 . cos 2x.✁

2 f ′(x) = a cos 2x √ sin 2x

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