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Chain Rule: Derivative of Composite Functions, Lecture notes of Calculus

The Chain Rule of differentiation, which allows finding the derivative of a composite function. The rule states that if g and f are differentiable functions, then the composite function F = f ◦ g is differentiable at x and its derivative F ′(x) is given by the product of f ′(g(x)) and g ′(x). examples to illustrate the application of the Chain Rule, including the Power Rule and the Quotient Rule.

What you will learn

  • What is the formula for the derivative of a composite function according to the Chain Rule?
  • How to find the derivative of a composite function using the Chain Rule?
  • Can the Chain Rule be applied to functions other than differentiable functions?

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2020/2021

Uploaded on 03/30/2021

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Lecture 10 : Chain Rule
Here we apply the derivative to composite functions. We get the following rule of differentiation:
The Chain Rule : If g is a differentiable function at x and f is differentiable at g(x), then the
composite function F = f g defined by F (x) = f(g(x)) is differentiable at x and F
0 is given by the
product
F0(x) = f0(g(x))g0(x).
In Leibniz notation If y=f(u) and u=g(x) are both differentiable functions, then
dy
dx =dy
du
du
dx.
It is not difficult to se why this is true, if we examine the average change in the value of F(x) that
results from a small change in the value of x:
F(x+h)F(x)
h=f(g(x+h)) f(g(x))
h=f(g(x+h)) f(g(x))
g(x+h)g(x)·g(x+h)g(x)
h
or if we let u=g(x) and y=F(x) = f(u), then
y
x=y
u·u
x
if g(x+h)g(x) = u6= 0. When we take the limit as h0 or x0, we get
F0(x) = f0(g(x))g0(x)
or dy
dx =dy
du
du
dx.
Example Find the derivative of F(x) = sin(2x+ 1).
Step 1: Write F(x) as F(x) = f(g(x)) or y=F(x) = f(u), where u=g(x).
Step 2: working from the outside in, we get
F0(x) = f0(g(x))g0(x) =
or using u, we get
F0(x) = dy
dx =dy
du ·du
dx .
Example Let g(x) = p(x3+x2+ 1)3, Find h0(x).
1
pf3
pf4
pf5

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Lecture 10 : Chain Rule

Here we apply the derivative to composite functions. We get the following rule of differentiation: The Chain Rule : If g is a differentiable function at x and f is differentiable at g(x), then the composite function F = f ◦ g defined by F (x) = f (g(x)) is differentiable at x and F ′^ is given by the product F ′(x) = f ′(g(x))g′(x).

In Leibniz notation If y = f (u) and u = g(x) are both differentiable functions, then

dy dx

dy du

du dx

It is not difficult to se why this is true, if we examine the average change in the value of F (x) that results from a small change in the value of x:

F (x + h) − F (x) h

f (g(x + h)) − f (g(x)) h

f (g(x + h)) − f (g(x)) g(x + h) − g(x)

g(x + h) − g(x) h

or if we let u = g(x) and y = F (x) = f (u), then

∆y ∆x

∆y ∆u

∆u ∆x

if g(x + h) − g(x) = ∆u 6 = 0. When we take the limit as h → 0 or ∆x → 0, we get

F ′(x) = f ′(g(x))g′(x)

or dy dx

dy du

du dx

Example Find the derivative of F (x) = sin(2x + 1).

Step 1: Write F (x) as F (x) = f (g(x)) or y = F (x) = f (u), where u = g(x).

Step 2: working from the outside in, we get F ′(x) = f ′(g(x))g′(x) = or using u, we get F ′(x) = (^) dxdy = dydu · dudx.

Example Let g(x) =

(x^3 + x^2 + 1)^3 , Find h′(x).

There is a general pattern with differentiating a power of a function that we can single out as:

The Chain Rule and Power Rule combined: If n is any real number and u = g(x) is differentiable, then d dx (un) = nun−^1 du dx or d dx ((g(x))n) = n(g(x))n−^1 g′(x).

Example Differentiate the following function:

f 1 (x) = sin^100 x.

We can combine the chain rule with the other rules of differentiation:

Example Differentiate h(x) = (x + 1)^2 sin x.

Example Find the derivative of the function

k(x) =

(x^3 + 1)^100 x^2 + 2x + 5

More Examples

Example(Old Exam Question Fall 2007) Find the derivative of

h(x) = x^2 cos(

x^3 − 1 + 2).

Example Find the derivative of

F (x) =

x^2 + x + 1

F (x) =

x^2 + x + 1

= (x^2 + x + 1)−^1 /^2.

By the chain rule,

F ′(x) =

(x^2 + x + 1)−^3 /^2 (2x + 1) =

−(2x + 1) 2(x^2 + x + 1)^3 /^2

Example Find the derivative of L(x) =

x− 1 x+.

Here we use the chain rule followed by the quotient rule. We have

L(x) =

x − 1 x + 2

x − 1 x + 2

Using the chain rule, we get

L′(x) =

x − 1 x + 2

d dx

x − 1 x + 2

Using the quotient rule for the derivative on the right, we get

L′(x) =

x − 1 x + 2

)− 1 / 2 [

(x + 2) − (x − 1) (x + 2)^2

]

x − 1 x + 2

)− 1 / 2 [

(x + 2)^2

]

Example f (x) =

cos^2 (2x)

Find f ′(0). (Note that this is an interesting function, in fact f (x) = | cos(2x)| which you can graph by sketching the graph of cos(2x) and then flipping the negative parts over the x-axis. Note that the graph has many sharp points, but is smooth at x = 0.)

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f ( x ) = cos ( 2 · x )

Using the chain rule with the chain y = f (x) =

cos^2 (2x) =

u, u = cos^2 (2x) = (v)^2 , v = cos(2x) = cos(w), w = 2x, we get

f ′(x) = dy dx

dy du

du dv

dv dw

dw dx

u−^1 /^2 · 2 v · [− sin(w)] · 2 =

1 2

cos^2 (2x)

· 2 cos(2x) · [− sin(2x)] · 2 = −2 cos(2x) sin(2x) √ cos^2 (2x)