We now have a pretty good list of “shortcuts” to find
derivatives of simple functions.
Of course, many of the functions that we will encounter
are not so simple. What is needed is a way to combine
derivative rules to evaluate more complicated functions.
®
Consider a simple composite function:
y = 6 x - 10
y = 2 (3x - 5)
If u = 3x - 5
then y = 2u
y = 6 x - 10
y = 2u
u = 3x - 5
dy
=6
dx
dy
=2
du
du
=3
dx
6 = 2×3
dy dy du
=
×
dx du dx
®
and another:
y = 5u - 2
where u = 3t
y = 5 (3t ) - 2 y = 5u - 2
u = 3t
y = 15t - 2
then y = 5 (3t ) - 2
dy
= 15
dt
dy
=5
du
du
=3
dt
15 = 5 × 3
dy dy du
=
×
dt du dt
®
and one more:
y = 9x + 6x +1
2
y = (3x + 1)
y = 9x + 6x +1
y =u
dy
= 18 x + 6
dx
dy
= 2u
du
2
2
u = 3x + 1
2
If u = 3x + 1
then y = u 2
This pattern is called
the chain rule.
du
=3
dx
dy
= 2 (3x + 1)
du
dy
= 6x + 2
du
18x + 6 = (6 x + 2 )× 3
dy dy du
=
×
dx du dx
®
Chain Rule:
If f
then:
dy dy du
=
×
dx du dx
g is the composite of y = f (u )and u = g (x ) ,
¢
f
g
(
) = fat¢ u = g (x) × gat¢ x
example:
f (x ) = sin x
g (x ) = x 2 - 4
f ¢ (x ) = cos x
g ¢ (x ) = 2 x
Find:
(f
g )¢ at x = 2
g (2 ) = 4 - 4 = 0
f ¢ (0 )× g ¢ (2 )
cos (0 )× (2 × 2 )
1× 4 = 4
®
We could also do it this way:
f (g (x )) = sin (x2 - 4 )
y = sin (x - 4 )
2
y = sin u
dy
= cos u
du
u = x2 - 4
du
= 2x
dx
dy dy du
=
×
dx du dx
dy
= cos u × 2 x
dx
dy
= cos (x 2 - 4 )× 2 x
dx
dy
= cos (22 - 4 )× 2 × 2
dx
dy
= cos (0 )× 4
dx
dy
=4
dx
®
Here is a faster way to find the derivative:
y = sin (x 2 - 4 )
d 2
y¢ = cos (x - 4 )× (x - 4 ) Differentiate the outside function...
dx
2
y¢ = cos (x 2 - 4 )× 2 x
…then the inside function
At x = 2, y¢ = 4
®
Another example:
d
cos 2 (3x )
dx
2
d
éëcos (3x )ùû
dx
d
2 éëcos (3x )ùû × cos (3x )
dx
It looks like we need to
use the chain rule again!
derivative of the
outside function
derivative of the
inside function
®
Another example:
d
cos 2 (3x )
dx
2
d
éëcos (3x )ùû
dx
d
2 éëcos (3x )ùû × cos (3x )
dx
d
2 cos (3x )× - sin (3x )× (3x )
dx
-2cos (3x )× sin (3x )× 3
-6cos (3x )sin (3x )
The chain rule can be used
more than once.
(That’s what makes the
“chain” in the “chain rule”!)
®
Derivative formulas include the chain rule!
d n
n -1 du
u = nu
dx
dx
d
du
sin u = cos u
dx
dx
d
du
cos u = - sin u
dx
dx
d
du
2
tan u = sec u
dx
dx
etcetera…
The formulas on the memorization sheet are written with
instead of du . Don’t forget to include the u ¢ term!
dx
u¢
®
The most common mistake on the chapter 3 test is to
forget to use the chain rule.
Every derivative problem could be thought of as a
chain-rule problem:
d 2
d
x = 2 x x = 2 x ×1 = 2x
dx
dx
derivative of
outside function
derivative of
inside function
The derivative of x is one.
®
The chain rule enables us to find the slope of
parametrically defined curves:
dy dy dx
= ×
dt dx dt
dy
dt = dy
dx
dx
dt
dx
Divide
sides
Theboth
slope
of aby
parametrized
curve is given by: dt
dy
dy
= dt
dx
dx
dt
®
Example:
x = 3cos t
y = 2sin t
These are the equations for
an ellipse.
dx
dy
= -3sin t
= 2 cos t
dt
dt
dy 2 cos t
2
=
= - cot t
dx -3sin t
3
Don’t forget to use the chain rule!
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