View Full Version : Chain rule from first principles examples?
wikid
Oct 12, 2011, 08:38 AM
"Differentiate y= square root(x^2 1) from fist principles." I have no idea where to begin. Please help!
ebaines
Oct 12, 2011, 09:19 AM
Some of your math symbols didn't come through. I assume you mean this:
y = \sqrt {x^2-1}
I won't solve this for you, but will give you another example that illustrates the process. You can then apply the process to your problem.
Suppose you want to differentiate
y = (x^3 +x^2)^2
You can start by substituting a new function g(x) for the part inside the parentheses:
g(x) = x^3 + x^2 \\
y = g(x)^2 \\
\frac {dy} {dg} = 2g(x)\\
\frac {dg} {dx} = 3x^2 +2x
Now apply the chain rule:
\frac {dy} {dx} = \frac {dy} {dg} \frac {dg} {dx} = 2g(x)(3x^2 + 2x) = 2(x^3+x^2)(3x^2+2x)
Hope this helps.
wikid
Oct 28, 2011, 07:17 AM
Yes, sorry, my symbols didn't really come through quite as I expected. I do understand how to differentiate a problem using the chain rule, which I assume is what you used in your example; however I am having trouble doing the same thing from first principles (you know, this one: http://intmstat.com/differentiation/Image536.gif) Thank you for helping though. By the way, you were right about your assumption of what I meant.
ebaines
Oct 28, 2011, 08:24 AM
The definition of the derivative is:
\frac {df } {dx} = \lim_{h \to 0} \frac { f(x+h)-f(x)}{h}
Using the chain rule:
\frac {df} {dx} = \frac {df}{dg} \cdot \frac {dg} {dx}
So let's take a look at the example I gave earlier. We can set:
g(x) = x^3 + x^2
and
f(g) = g^2
Now we find the derivatives of these two functions from first principals.
\frac {dg } {dx} = \lim_{h \to 0} \frac { ((x+h)^3+(x+h)^2)-(x^3+x^2)}{h}
= \lim_{h \to 0} \frac {( x^3+3x^2h+3xh^2+h^3 +x^2 +2xh+h^2)- (x^3+x^2)}{h} \\
= \lim_{h \to 0} \frac {3x^2h+3xh^2+h^3+2xh + h^2}{h}
= \lim_{h \to 0} ( 3x^2+3xh+h^2+2x+h)
= 3x^2 + 2x
and:
\frac {df } {dg} = \lim_{h \to 0} \frac { (g+h)^2-g^2)}{h}
= lim_{h \to 0} \frac {2gh+h^2}{h} = 2g
Then apply the chain rule:
\frac {df}{dx} = \frac {df}{dg} \frac {dg}{dx} = 2g(3x^2+2x) = 2(x^3 + x^2)(3x^2+2x)
You can use this same process on your problem, setting g(x) = x^2-1 and f(g) = \sqrt g. Give it a try and post back with what you get.
Hope this helps.