Why is the chain rule used

How do I apply the chain rule?

Derivation with the chain rule: application

The chain rule is one important rulewith the help of which you can derive complex $ _ "$ chained $" $ functions. The chain rule occurs primarily in combination with other rules, such as the factor rule or the sum rule. An example of the application of the chain rule is this function:

$ \ large {f (x) = (x ^ 2) ^ 3} $

We want to demonstrate the chain rule, so we don't add up the exponents. First we have to define the two functional parts u (x) and v (x), because the chain rule says that:

$ f (x) = \ textcolor {green} {u (} \ textcolor {blue} {v (x)} \ textcolor {green} {)} $

In our example this is Functional part $ \ textcolor {blue} {x ^ 2} $ the part $ \ textcolor {blue} {v (x)} $ and the Functional part $ \ textcolor {green} {() ^ 3} $ the part $ \ textcolor {green} {u (x)} $, so:

$ f (x) = \ textcolor {green} {(} \ textcolor {blue} {x ^ 2} \ textcolor {green} {) ^ 3} $

Now let's look at them again Derivation formula $ f '(x) = \ textcolor {green} {u' (} \ textcolor {blue} {v (x) \ textcolor {green} {)} \ cdot v '(x)} $. So we have to do the first Derivatives of the individual functional parts and then to calculate them Derivative function put together.

So we form them Derivatives the individual Functional parts:

  • $ \ textcolor {blue} {v '(x) = 2x} $, v (x) is also called inner function designated.
  • $ \ textcolor {green} {u '(x) = 3 \ cdot () ^ 2} $, u (x) is also called external function designated.

After this Put together we receive:

$ f '(x) = 3 \ cdot (x ^ 2) ^ 2 \ cdot 2x $

We can simplify this term and then get:

$ f '(x) = 6 \ cdot x ^ 5 $

Incidentally, we could have summarized the function first and then derived it with the help of the power rule. So:

$ {f (x) = (x ^ 2) ^ 3} ~~~ \ rightarrow ~~~ {f (x) = x ^ 6} $

$ {f '(x) = 6 \ cdot x ^ 5} $

Chain rule: example

AnotherSample task for the chain rule, in conjunction with the Sum rule, is the function:

$ g (x) = \ textcolor {green} {(} \ textcolor {blue} {3x-2} \ textcolor {green} {) ^ 8} $

The derivation of the external function is: $ \ textcolor {green} {u '(x) = 8 \ cdot () ^ 7} $

The derivation of the inner function is: $ \ textcolor {blue} {v '(x) = 3} $

Merged according to the Chain rule this gives the Derivation the function:

$ g '(x) = \ textcolor {green} {8 \ cdot (} \ textcolor {blue} {3x-2} \ textcolor {green} {) ^ 7} \ cdot 3 $

If we simplify this term, we get: $ g '(x) = 24 \ cdot (3x-2) ^ 7 $

To deepen the topic, take a look at the Exercisesto chain rules.