The Chain Rule Given a composite function: The Chain Rule Given a - PowerPoint PPT Presentation

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

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:

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 x 2

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: x 2 h ( x ) = sin ���� outer function

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 x 2

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: x 2 h ( x ) = sin ���� inner function

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 x 2

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 x 2 h ′ ( x ) = cos x 2 .

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 x 2 cos x 2 h ′ ( x ) = . � �� � derivative of outer function

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 x 2 h ′ ( x ) = cos x 2 .

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 x 2 h ′ ( x ) = cos x 2 . 2 x

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 x 2 h ′ ( x ) = cos x 2 . 2 x ���� derivative of inner function

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 x 2 h ′ ( x ) = cos x 2 . 2 x

Example 1

Example 1 √ f ( x ) = 3 sin x

Example 1 √ f ( x ) = 3 sin x We can write it as:

Example 1 √ f ( x ) = 3 sin x We can write it as: 1 f ( x ) = 3 (sin x ) 2

Example 1 √ f ( x ) = 3 sin x We can write it as: 1   2 f ( x ) = 3 sin x   ���� inner function

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ ✘✘ 2 f ( x ) = 3 sin x

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2 We take the derivative of the outer function:

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2 We take the derivative of the outer function: f ′ ( x ) = 3 . 1 2 y − 1 2 .

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2 We take the derivative of the outer function: f ′ ( x ) = 3 . 1 2 y − 1 2 . y ′

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2 We take the derivative of the outer function: f ′ ( x ) = 3 . 1 2 . y ′ = 3 2 y − 1 2 (sin x ) − 1 2 . (sin x ) ′

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2 We take the derivative of the outer function: f ′ ( x ) = 3 . 1 2 . y ′ = 3 2 . (sin x ) ′ = 3 2 y − 1 2 (sin x ) − 1 2 (sin x ) − 1 2 . cos x

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2 We take the derivative of the outer function: f ′ ( x ) = 3 . 1 2 . y ′ = 3 2 . (sin x ) ′ = 3 2 y − 1 2 (sin x ) − 1 2 (sin x ) − 1 2 . cos x f ′ ( x ) = 3 cos x √ sin x

Example 1 √ f ( x ) = 3 sin x We can write it as: � 1 � ✿ y ✘ 2 = 3 y ✘✘ 1 f ( x ) = 3 sin x 2 We take the derivative of the outer function: f ′ ( x ) = 3 . 1 2 . y ′ = 3 2 . (sin x ) ′ = 3 2 y − 1 2 (sin x ) − 1 2 (sin x ) − 1 2 . cos x f ′ ( x ) = 3 cos x √ sin x

Example 2

Example 2 √ f ( x ) = a sin 2 x

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2 f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 .

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2 f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . (sin 2 x ) ′

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2 f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . (sin 2 x ) ′ f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . cos 2 x .

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2 f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . (sin 2 x ) ′ f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . cos 2 x . 2

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2 f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . (sin 2 x ) ′ f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . cos 2 x . ✁ 2 ✁

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2 f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . (sin 2 x ) ′ f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . cos 2 x . ✁ 2 ✁ f ′ ( x ) = a cos 2 x √ sin 2 x

Example 2 1 √ f ( x ) = a sin 2 x = a (sin 2 x ) 2 f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . (sin 2 x ) ′ f ′ ( x ) = a . 1 2 . (sin 2 x ) − 1 2 . cos 2 x . ✁ 2 ✁ f ′ ( x ) = a cos 2 x √ sin 2 x