Transcript Section 3.4 The Chain Rule

Section 3.4
The Chain Rule
One of THE MOST POWERFUL
Rules of Differentiation
The chain rule allows you to take derivatives of
compositions of functions that may be hard or
even impossible to differentiate with previous
rules only.
Examples:


f ( x)  x  1

2

f ( x)  x  1
2
2
4
not much fun, but
doable with previous
techniques
Previously impossible
(Unless you want to try to definition)
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The Chain Rule
If y=f(u) is a differentiable function of u, and
u=g(x) is a differentiable function of x, then
y=f(g(x)) is a differentiable function of x and
 dy/dx = dy/du du/dx
OR
d
[ f ( g ( x)]  f ' ( g ( x)) g ' ( x)
dx
What does this mean?
When taking the derivative of a composite
function, you first take the derivative of the
outside function at the inner function and
then multiply by the derivative of the inner
function.
Common Types
1)One of the most common types of composite
functions is y  [u( x)]n
What the chain rule tells us to do in this case is take
the derivative with respect to the outside exponent,
and leave the inner function alone. Then, we multiply
by the derivative of the inner function (what was being
raised to the power)
Examples:
y  (9t  7)
8
y
3
x  3
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What about Trig functions and
Exponentials?
d
du
[sin u ]  (cos u )
dx
dx
d
du
2
[tan u ]  (sec u )
dx
dx
d
du
[sec u ]  (sec u tan)
dx
dx
d u
u du
[e ]  (e )
dx
dx
d
du
[cos u ]  (sin u )
dx
dx
d
du
2
[cot u ]  (csc u )
dx
dx
d
du
[csc u ]  (csc u cot u )
dx
dx
Examples
y  sin
x  sin x
y  2 tan x
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Exponentials and Logarithms
Let a be a positive real number not = 1 and
let u be a differentiable function of x.
d
1 du
d
1
[ln x]  , x  0 dx [ln u ]  y dx , u  0
dx
x
 
d x
x
a  (ln a)( a )
dx
d u
u du
a  (ln a)( a )
dx
dx
 
d
1 du
[ln u ] 
dx
u dx
d
1
log a x 
dx
(ln a) x
d
1 du
log a u  
dx
(ln a)u dx
More Examples:
y  ln( x  2)
3
y  sec x
y6
3x
3
And more examples…
The chain rule can be used in conjunction with both
product and quotient rules. You will need to decide
what you should do first.
 2x  5 
g ( x)   2

 x 2
2 x  5
5
g ( x) 
x2  2
5