"Differentiate y= square root(x^2 1) from fist principles." I have no idea where to begin. Please help!

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.

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.

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.