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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)
