[Blindmath] Chain Rule

[Joseph Lee]

Hi,
Let's see:
Are ^ ast1liar with a concept of "composiTion of functions" where 
a function becomes input to another function? If you do, then 
it'll be easier to understand how chain rule works; if not, I 
suggest studyounging that concept before moving onto chain rule 
(but I'm sure you know it already).
The chain rule works like this: If I have composi;name of 
func;arns such as f of g of x, then the derivative would be 
derivative of g of f times derivative of f, written as:
y = f(gx)x; dy/dx = d(g)f*df.
For example, suppose I have f being x^4 and g being 3x.  The 
steps are:
1.  Find the derivative of g (inner function) first: dy/dx of 3x 
= 3.
2.  Find the derivative of f (outer function): dy/dx of x^4 = 
4x^3.
3.  Plug in g to the derivative of f, replacing any variables (x 
or whatever) with the original g: dy/dx 4(g)^3 becomes 4(3x)^3.
4.  Multiply the third expression above with derivative of g: 
dy/dx = 3*(4(3x)^3).  We can simlify this a bit further to get 
the final answer.
Thus: y = (3x)^4; dy/dx = 3*4*(3x)^3 = 12(3x)^3.
There are some good tutorials on YouTube on how to use variable 
derivative rules (including chain rule), which are easy to 
follow.  To get up to speed with chain rule, do some practice 
examples.
Hope this helps.
Cheers,
Joseph


 ----- Original Message -----
From: "Duong Tuan Nam" <tuannamduong at gmail.com
To: "Blind Math list for those interested in 
mathematics"<blindmath at nfbnet.org
Date sent: Fri, 4 May 2012 23:36:11 +0700
Subject: [Blindmath] Chain Rule

Hi all,
Could anyone help me explain how chain rule works? I try to read 
on Wikipedia but I've still not actually comprehended the 
concept.

I much appreciate your help.

Have a nice day,
Nam
Duong Tuan Nam
Email: tuannamduong at gmail.com
Yahoo ID: tuannamfriend
Skype Name: duongtuannam1992
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