First Principles

[mcb30]

Hi! I have an assignment to find the derivatives of two funtions (1/t and the square root of t). I know how to find derivatives, (a couple of ways, actually) I've just never been taught something with the name 'the first principle of calculus.' How does he want me to find them? I probably already know the method, just not by that name.

 

I saw Dave's lessons, but for some reason they don't display properly on my computer. ( portions of the text are bold and missing) Sorry to ask for something I know is already posted, but I can't get it to display properly.

 

Thanks!

Maria

[matt grime]

if f is your fucntion, then the derivative at x is the limit of

 

(f(x+h)-f(x))/h

 

as h tends to zero.

 

example: prove that the derivative of x^2 is 2x:

 

we need to find the limit of

 

((x+h)^2-x^2)/h

 

ok, well, let's expand it and and cancel off the h as we may and we need to find the limit as h tends to zero of

 

2x+h

 

 

which is just 2x as requried

 

 

no, try if for (sqrt(x+h)-sqrt(x))/h

 

it will help to bear in mind that s^2+t^2=(s-t)(s+t) (or setting sqrt(u)=s and sqrt(v)=t) may be more illuminating) so try multiplying that expression top and bottom by the same thing

 

now try simplifying for {1/(t+h) - 1/t}/h as well and let the h's tend to zero

[mcb30]

Thanks!

 

Sorry I couldn't have just used the thread that was already posted!

I have worked with that method before, I've just never heard it called by that name.

 

That clears everything up!

 

Maria

[mezarashi]

If memory serves, by first principles means by the original way that your rules were derived. You know the power rule, the sine, cosine, tangent, log, exponential rules right? Who made these rules up? The answer: they were derived from first principles and for integrals many from the fundamental theorem of calculus.

 

The first principles method is to use the very basic definition, that is, the derivative of any point is the change in y, delta y, divide by the change in x, delta x, all of this as delta y and delta x approach zero. What you get is an equation that looks like this:

 

Take for example your function is sin x:

 

[math]

 

\frac{dy}{dx} = \lim_{\Delta x\to0}\frac{\sin{( x+\Delta x )}-\sin{x}}{\Delta x}

 

[/math]

 

Pheew, that took me forever to spit out in latex. Hardsss x_x

Edit: yeap, long it was