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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:

) 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.