Calculus I
Notes on The Chain Rule.
Example #1: Find the derivative of f (x)   5x 4  7  .
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f (x)   5x 4  7    5x 4  7    5x 4  7  
f '(x)   20x 3    5x 4  7    5x 4  7     5x 4  7   dxd  5x 4  7    5x 4  7 
  20x 3    5x 4  7    5x 4  7     5x 4  7     20x 3    5x 4  7    5x 4  7    20x 3  
  20x 3    5x 4  7    5x 4  7    5x 4  7    20x 3    5x 4  7    5x 4  7    5x 4  7    20x 3 
Let’s notice at this point that we could expand the product rule for derivatives to 3 factors to
d
f (x)  g(x)  h(x)  f '(x)  g(x)  h(x)  f (x)  g '(x)  h(x)  f (x)  g(x)  h '(x)
dx 
Now getting back the problem, let’s condense the derivative.
f '(x)  3   5x 4  7   20x 3   60x 3  5x 4  7 
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2
2
Look at the function from the previous example and the first 2 factors of its derivative in the second to last form.
f (x)   5x 4  7 
3
f '(x)  3   5x 4  7    20x 3 
2
Note that we in effect we did the derivative of the cube just like we did before.
d
dx
 x 3   3x 2 . We just left what was inside the cube
alone. Now notice that what is in the brackets, f '(x)  3   5x 4  7    20x 3  , is the derivative of what is inside the cube. This is the
chain rule.
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The Chain Rule
Informal:
To take the derivative of a function within a function, first take the derivative of the outermost function leaving what’s inside alone.
Then multiply by the derivative of the inner function.
Formal:
For all x where g(x) is differentiable at x and f (u) is differentiable at u  g(x) , then
d
dx
or for u  g(x) and y  f (u) ,
f  g(x)   f '  g(x)   g '(x)
dy dy du


dx du dx
Proof:
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d
dx
Lim
f  g (x)   
x  0

Lim
x  0

Lim
u  0
 f  g (x  x)   f  g (x)  


x


 f  g (x  x)   f  g (x)  g (x  x)  g (x)  substitute u  g(x) and u  g (x  x)  g (x)


,
and note that as x  0, u  0
g (x  x)  g (x)
x


 f  u  u   f  u  
Lim
 g (x  x)  g (x) 

 


u
x


 x  0 
 f '  u   g '(x)  f '  g (x)   g '(x)
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Seminole State:Rickman
Notes on The Chain Rule.
Page #1 of 2
Example #2: Find the derivatives of
a) y   7x 2  3x  2 
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b) y  sin(6x  3)
c) y  cos 2 (x)
e) y  csc  x 8 
d) y  tan 6 (x)
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Remember that with all of these, you need to be concerned with order of operations, and you deal with the outer function first. Then
multiply by the derivative of what’s inside.
a) y '  5  7x 2  3x  2   14x  3
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b) y'  cos(6x  3)  6  6cos(6x  3)
c) y  cos2 (x)   cos(x)  . Thus, y '  2  cos(x)   -sin(x)  -2cos(x) sin(x)
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2
d) y  tan 6 (x)   tan(x)  . Hence, y '  6  tan(x)   sec2 (x)   6 tan 5 (x)sec2 (x)
e) y '  -csc  x8  cot  x8  8x 7   -8x 7 csc  x8  cot  x8 
5
6
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Example #3: Find the equation of the tangent line to
f (x)  csc8 (x) at x  34
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y  csc8  34    2   16
8
m  f '(x) x  3  8csc7 (x) -csc(x) cot(x) 
4
x  34
 -8csc8 (x) cot(x)
x  34
 -8csc8  34  cot  34   -8
 2  1  -128
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y  8  -128  x  34 
y  8  -128x  96
y  128x  96  8
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Now that we have the chain rule, we can finally do derivatives of functions with trigonometric with more than just x in a
trigonometric function. Thus, we can do more realistic functions.
Example #4: For a AC circuit the function for the voltage is given by V(t) 
2
2
VRMS sin(2 f  t) , where t is in seconds and V(t) is
cycles
given in volts. Also for a standard wall AC outlet, VRMS  110volts , and the frequency, frequency  f  60 second
. Find the rate of
change in the voltage of a standard wall AC outlet at t  4.85sec. Round the answer to the thousandths.
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V(t)  22 110  sin(2  60  t)  77.8sin(120  t)
rate of change  dtd  V(t)   77.8cos(120  t) 120  9336 cos(120  t)
at t  4.85sec.
rate of change  9336 cos(120  4.85)  9336 cos(1828.4)  29329.909
Since V(t) is in volts and t is in seconds, the rate of change would be in
volts
second
rateof change  29329.909
. Finally,
volts
second
.
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Example #5: An inductor is an electronic component that when the current is changing it creates a voltage in opposition to the
circuit’s voltage. This induced voltage, VL , in volts is given by VL  - L dIdt where L is the inductance measured in Henries, H, and I
is current measured in amperes, A, and t is time in seconds. For an AC circuit with an inductor, if I 10sin(120 t) and L  0.02H ,
find the maximum induced voltage.
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dI
10cos(120 t)  120 1200 cos(120 t)
dt
VL  -0.02 1200 cos(120 t)
 - 24 cos(120 t)
Thus, the max induced voltage would be 24 V  75.40V .
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Seminole State:Rickman
Notes on The Chain Rule.
Page #2 of 2