[Blindmath] Chain Rule

[Duong Tuan Nam]

Thank you, Joseph,
I've comprehended.
Nam
----- Original Message ----- 
From: "Joseph Lee" <joseph.lee22590 at gmail.com>
To: "Blind Math list for those interested in mathematics" 
<blindmath at nfbnet.org>
Sent: Saturday, May 05, 2012 3:19 AM
Subject: Re: [Blindmath] Chain Rule


> 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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