There are rules we can follow to find many derivatives.
If we know the rate of change for two related things, how do we work out the overall rate of change?
The Chain Rule tells us how!
Example: Sage the Dog can run 3 times faster than you, and you can run 2 times faster than me, so Sage can run 3 × 2 = 6 times faster than me.
Let's use some notation. Call the dog "y", me "x" and you can be "u":
dy dx = dy du du dx
dy dx = dy du du dx = 3 × 2 = 6
But it is not usually that easy!
Because one function can depend on the current value of the other (which is itself continually changing).
There are two functions happening here, sin() and x 2 .
But it is not sin(x), it is sin(the result of x 2 )
Let's use "u" for x 2 so we can have:
dy dx = dy du du dx
d dx sin(x 2 ) = d du sin(u) d dx x 2
The individual derivatives are:
d dx sin(x 2 ) = cos(u) (2x)
Substitute back u = x 2 :
d dx sin(x 2 ) = cos(x 2 ) (2x)
Which is neater this way:
d dx sin(x 2 ) = 2x cos(x 2 )
There are several different notations that can be used!
Notation | Chain Rule | |
---|---|---|
Using d dx | dy dx = dy du du dx | |
Using ’ (meaning derivative of) | f(g(x)) = f’(g(x))g’(x) | |
As "Composition of Functions" | f º g = (f’ º g) × g’ |
Let's do the previous example again using f(g(x)) = f'(g(x))g'(x):
sin(x 2 ) is made up of sin() and x 2 :
The Chain Rule says:
the derivative of f(g(x)) = f'(g(x))g'(x)
The individual derivatives are:
d dx sin(x 2 ) = cos(g(x)) (2x)
Same result as before (thank goodness!)
Another couple of examples:
1/cos(x) is made up of 1/g and cos():
The Chain Rule says:
the derivative of f(g(x)) = f’(g(x))g’(x)
The individual derivatives are:
(1/cos(x))’ = −1g(x) 2 (−sin(x))
= sin(x)cos 2 (x)
Note: sin(x)cos 2 (x) is also tan(x)cos(x) or many other forms.
The Chain Rule says:
the derivative of f(g(x)) = f’(g(x))g’(x)
(5x−2) 3 is made up of g 3 and 5x−2:
The individual derivatives are:
ddx (5x−2) 3 = (3g(x) 2 )(5)