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Derivative of Composite Functions: The Chain Rule, Study notes of Calculus

The concept of derivative of composite functions using the chain rule. It provides examples and formulas to help understand the product of derivatives evaluated appropriately for different functions. Students of calculus and mathematics can benefit from this document as study notes, summaries, or cheat sheets for understanding the chain rule.

Typology: Study notes

2021/2022

Uploaded on 09/27/2022

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Derivatives of Composite Functions
Recall that the composite function or composition of two functions is the function
obtained by applying them one after the other.
For example, If f(x) =1
xand g(x) =x3+2, then
f(g(x))=1
g(x) =1
x3+2
and g(f(x)) =(f(x)
)3+2=1
x3
+2=1
x3+2
Try a Java applet.
The derivative of the composition of two non-constant functions is equal to the product
of their derivatives, evaluated appropriately.
The Chain Rule
We have the Chain Rule: g(h(x))=g(h(x))h(x)
Example 1: Using g(x) =1
x=x1and h(x) =x3+2,
we have g(x) =(1)x2and h(x) =3x2,g(h(x)) =(1)(h(x))2, soweget
1
x3+2=g(h(x))h(x) =(1)(h(x))2(3x2)=
(1)(x3+2)2(3x2)=3x2
(x3+2)2
On the other hand, 1
x3+2=h(g(x))=h(g (x))g(x) =
3(g(x))2(x2)=3(x 1)2(x2)= 3x4, as expected.
1
pf3
pf4

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Derivatives of Composite Functions

Recall that the composite function or composition of two functions is the function

obtained by applying them one after the other.

For example, If f (x) =

x

and g(x) = x^3 + 2, then

f (g(x)) =

g(x)

x^3 + 2

and g(f (x)) = (^) (f (x))

3

  • 2 =

x

x^3

Try a Java applet.

The derivative of the composition of two non-constant functions is equal to the product

of their derivatives, evaluated appropriately.

The Chain Rule

We have the Chain Rule:

g(h(x))

= g

(h(x))h

(x)

Example 1: Using g(x) =

x

= x

− 1 and h(x) = x^3 + 2,

we have g(x) = ( − 1 )x −^2 and h(x) = 3 x^2 , g(h(x)) = ( − 1 )(h(x)) −^2 , so we get

x^3 + 2

= g

(h(x))h

(x) = ( − 1 )(h(x))

− 2 ( 3 x

2 ) =

( − 1 )(x

3

  • 2 )

− 2 ( 3 x

2 ) =

− 3 x^2

(x^3 + 2 )^2

On the other hand,

x^3

h(g(x))

= h

(g(x))g

(x) =

3 (g(x))

2 (x

− 2 ) = 3 (x

− 1 )

2 (x

− 2 ) = − 3 x

− 4 , as expected.

Example 2: Let g(x) = x^3 , and h(x) = x^2 , so that

g(h(x)) = h(x)^3 = (x^2 )^3 = x^6.

Then g(x) = 3 x^2 , so g(h(x)) = 3 (h(x))^2 , and h(x) = 2 x ,

so the Chain Rule gives us

g(h(x))

= g(h(x))h(x) =

3 (h(x))^2

( 2 x) =

3 (x^2 )^2

( 2 x) =

3 x^4

( 2 x) = 6 x^5 , as expected.

Example 3: Let g(x) = x^3 + 3, and h(x) = x^2 + 2, so that

g(h(x)) = (h(x))^3 + 3 = (x^2 + 2 )^3 + 3.

Then g(x) = 3 x^2 , so g(h(x)) = 3 (h(x))^2 , and h(x) = 2 x ,

so the Chain Rule gives us

g(h(x))

= g(h(x))h(x) =

3 (h(x))^2

( 2 x) =

3 (x^2 + 2 )^2

( 2 x) =

6 x(x

2

  • 2 )

2

Example 4: Find f ′ (x) if f (x) =

x^4 + x^2 + 1.

We let g(x) = x

1 (^3) and h(x) = x^4 + x^2 + 1 so that f (x) = g(h(x)).

Then g(x) =

1 3 x

− (^23) , g(h(x)) = 1 3

(h(x))

− (^23) , and h(x) = 4 x^3 + 2 x ,

so we have f(x) = g(h(x))h(x) =

(h(x))

− (^23) ( 4 x^3 + 2 x) =

2 x( 2 x

2

  • 1 )

3 (x^4 + x^2 + 1 )

2 3

Example 8: Find f ′ (x) if f (x) = e

cos x .

We have f

(x) = e

cos x ( cos x)

′ = e

cos x ( − sin x) = − sin xe cos^ x

Example 9: Find f ′ (x) if f (x) = sin

e

tan x ) .

We have f

(x) = cos

e

tan x ) ( e

tan x )′^ = cos

e

tan x )^ e

tan x ( tan x)

cos

e

tan x )^ e

tan x sec

2 x