Question 1203098
<pre>
I taught calculus for 40 years but never heard of "first principle".  So I
assumed you meant the first derivative formula you learn which is the derivative
of a power of x.  You haven't gotten to the short-cut formulas yet.

What you want is what we called "by using the definition of the derivative" or
"by the limit of the difference quotient".

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(y(x+h)-y(x) )/h,
               h->0,"")}}}

-------------------------------------------------

{{{y}}}{{{""=""}}}{{{3x - 1/(2x)}}}

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

( (3(x+h) - 1/(2(x+h))) - (3x - 1/(2x) ) )/h^"",
               h->0,"")}}}

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

( 3x+3h - 1/(2(x+h)) - 3x + 1/(2x)  )/h^"",
               h->0,"")}}} 

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(   3h - 1/(2(x+h)) + 1/(2x)  )/h^"",
               h->0,"")}}}

Multiply top and bottom by LCD = 2x(x+h)

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(   6hx(x+h) - x + (x+h)  )/(2hx(x+h)),
               h->0,"")}}}

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(   6hx^2+6h^2x - x + x+h  )/(2hx(x+h)),
               h->0,"")}}}

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(   6hx^2+6h^2x+h  )/(2hx(x+h)),
               h->0,"")}}}

Factor h out of the numerator:

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(h(6x^2+6hx+1))/(2hx(x+h)),
               h->0,"")}}}

Cancel h's

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(cross(h)(6x^2+6hx+1))/(2*cross(h)x(x+h)),
               h->0,"")}}}

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{matrix(2,2, lim,

(6x^2+6hx+1)/(2x(x+h)),
               h->0,"")}}}

As h approaches 0 the numerator and denominator approach what we
get if we substitute 0 for h, so the limit as h approaches 0 is:

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{(6x^2+6(0)x+1)/(2x(x+(0)))}}}

{{{dy/dx}}}{{{""=""}}}{{{"y'"}}}{{{""=""}}}{{{(6x^2+1)/(2x^2)}}}

Edwin</pre>