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Chain Rule: Derivative of Composite Functions, Study notes of Applied Differential Equations

The chain rule is a fundamental concept in calculus that allows finding the derivative of a composite function. This rule states that the derivative of a function that is a composition of another function and a constant or variable function is the product of the derivatives of each function. examples and explanations of how to apply the chain rule to find derivatives of composite functions.

What you will learn

  • Can the chain rule be applied to any type of composite function?
  • What is the formula for the chain rule?
  • How does the chain rule work for finding the derivative of a composite function?

Typology: Study notes

2021/2022

Uploaded on 09/27/2022

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The Chain Rule:
d
dx(f(g(x)) = f0(g(x)) ·g0(x).
Example 1. Find the derivative of y=x3+x+ 2 = f(g(x)), where
f(u) = u=u1/2and g(x) = x3+x+ 2, so...
d
dxpx3+x+ 2 = d
dx x3+x+ 21/2=
f0(g(x))
z }| {
1
2x3+x+ 21/2(3x2+ 1)
| {z }
g0(x)
(*) The chain rule can also be expressed as follows. If y=f(u) and
u=g(x), then y=f(g(x)) and
dy
dx =
f0(g(x))
z}|{
dy
du ·
g0(x)
z}|{
du
dx
1
pf3
pf4
pf5
pf8
pf9
pfa

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Download Chain Rule: Derivative of Composite Functions and more Study notes Applied Differential Equations in PDF only on Docsity!

The Chain Rule:

d dx (f (g(x)) = f ′(g(x)) · g′(x).

Example 1. Find the derivative of y =

x^3 + x + 2 = f (g(x)), where

f (u) =

u = u^1 /^2 and g(x) = x^3 + x + 2, so...

d dx

x^3 + x + 2 = d dx

x^3 + x + 2

f ′(g(x)) ︷ ︸︸ ︷ 1 2

x^3 + x + 2

(3x^2 + 1) ︸ ︷︷ ︸ g′(x)

(*) The chain rule can also be expressed as follows. If y = f (u) and u = g(x), then y = f (g(x)) and

dy dx

f ′(g(x)) ︷︸︸︷ dy du

g′(x) ︷︸︸︷ du dx

Explanation: With y = f (u) and u = g(x), linear approximation says that if ∆x ≈ 0, then

∆u = g(x + ∆x) − g(x) ≈ g′(x)∆x. (1)

Likewise, if ∆u ≈ 0, then

∆y = f (u + ∆u) − f (u) ≈ f ′(u)∆u. (2)

Now, if ∆x is close to 0, then so is ∆u (because g′(x) is fixed), so if ∆x ≈ 0, then using both approximations (1) and (2) gives

∆y ≈ f ′(u)∆u ≈ f ′(u)

︷ ≈︸︸∆u ︷ g′(x)∆x = f ′(g(x))g′(x)∆x,

so ∆y ∆x

f ′(g(x))g′(x)∆x ∆x^

= f ′(g(x))g′(x).

These approximations all becomes more and more accurate as ∆x → 0, and therefore dy dx = (^) ∆limx→ 0 ∆y ∆x = f ′(g(x))g′(x).

-3 -2 -1 0 1 2 3 4 5 6

4

8

12

y=(x^2 -3x-4)^3

Figure 1: The graph of y = (x^2 − 3 x − 4)^3.

Differentiating without the chain rule...

h(x) = (x^2 − 3 x − 4)^3 = (x^2 − 3 x − 4)^2 (x^2 − 3 x − 4) = (x^4 − 6 x^3 + x^2 + 24x + 16)(x^2 − 3 x − 4) = x^6 − 9 x^5 + 15x^4 + 45x^3 − 60 x^2 − 144 x − 64

So

h′(x) = 6x^5 − 45 x^4 + 60x^3 + 135x^2 − 120 x − 144.

Now all we have to do is to solve the equation

6 x^5 − 45 x^4 + 60x^3 + 135x^2 − 120 x − 144 = 0...

Observation: Using the chain rule in this example has two advantages:

(*) No messy arithmetic.

(*) The chain rule gives h′(x) in a (partially) factored form, which makes solving the equation h′(x) = 0 is much easier.

-2 -1 0 1 2 3 4 5

1

2

3

y=2(x^2 +4)-1/

y=4/3-x/

Figure 2: The graphs of y = 2(x^2 + 4)−^1 /^3 and the tangent line at (2, 1).

Observation: f (x)/g(x) = f (x)g(x)−^1 , so...

d dx

f (x) g(x)

d dx

f (x)g(x)−^1

= f ′(x)g(x)−^1 + f (x) d dx

g(x)−^1

= f ′(x)g(x)−^1 + f (x)

(−1)g(x)−^2 g′(x)

f ′(x) g(x)

f (x)g′(x) g(x)^2

f ′(x)g(x) − f (x)g′(x) g(x)^2

I.e., the quotient rule follows from combining the product rule and chain

rule.

Example: The demand equation for a firm’s product is

p = 100 − 0. 8 q

and the firm’s production function is

q = 5

4 l − 15 ,

where labor input l is measured in 40-hour work-weeks.

(*) Find the firm’s marginal revenue product when l = 10.

  1. r = pq = 100q − 0. 8 q^2 =⇒ dr dq = 100 − 1. 6 q

dq dl

d dl

5(4l − 15)^1 /^2

= 5 · 12 (4l − 15)−^1 /^2 · 4 = 10(4l − 15)−^1 /^2

  1. q(10) = 5

dr dl

l=

dr dq

q=

×

dq dl

l=

︷ ︸︸^60 ︷

(100 − 1. 6 · 25) ×

︷ ︸︸^2 ︷

10 · 25 −^1 /^2