Notation: "delta" is denoted here by "ch" (for "change")

Rule 1: Rule of constants: Y=b => dY/dX = 0
    If Y = b, then: (1) Y = f(X) = b and (2) Y+chY = f(X+chX) = b. So:
        chY   f(X+chX) - f(X)   b - b    0
        --- = --------------- = ----- = --- = 0. QED
        chX         chX          chX    chX

Rule 2: Power rule: Y = Xⁿ => dY/dX = nX^n-1

    This is proven last, because it is the longest proof, and part of it relies on the quotient rule and
the chain rule.

Rule 3: Rule of coefficients: Y = b u(X) => dY/dX = b du/dX
    If Y = bu(X), then Y+chY = b(u+ch(u)). So:

        chY   b(u+ch(u)) - bu   bu + b ch(u) - bu    ch(u)
        --- = --------------- = ----------------- = b-----
        chX         chX               chX             chX

        dY            ch(U)     du
        -- = lim     b-----  = b--. QED
        dX    chX->0    chX      dx

Rule 4: Summation rule: Y = u + v => dY/dX = du/dX + dv/dX
    If Y = u + v, then Y+chY = (u+ch(u)) + (v+ch(v)). So:

        chY   (u + ch(u)) + (v + ch(v)) - (u + v)   ch(u)   ch(v)
        --- = ----------------------------------- = ----- + -----
        chX                 chX                      chX     chX

        dY          ch(u)   ch(v)   du   dv
        -- = lim    ----- + ----- = -- + --.   QED
        dX   chX->0  chX     chX    dX   dX

Rule 5: Product rule: Y = u v => dY/dX = u dv/dX + v du/dX
    If Y = uv, thenY+chY = (u+chu)(v+chv) = uv + u chv + chu v + chu chv. So:

        chY   (uv + u chv + chu v + chu chv) - uv    chv    chu      chv
        --- = ----------------------------------- = u--- + v--- + chu---
        chX                 chX                      chX    chX      chX

        dY           chv    chu      chv     dv    du      dv    dv    du
        -- = lim   (u--- + v--- + chu---) = u-- + v-- + chu-- = u-- + v--
        dX   chX->0  chX    chX      chX     dx    dx      dx    dx    dx

                since chu-> 0 as chX-> 0. QED
 
                                                                    v du/dX - u dv/dX
Rule 6: Quotient rule: Y = u/v => dY/dX =  ----------------------
                                                                              v²
    Since Y = u/v, Y+chY = (u+chu)/(v+chv). So:

              u+chu   u   (u+chu)v - u(v+chv)   v chu - u chv
        chY = ----- - - = ------------------- = -------------
              v+chv   v        v(v+chv)           v² + v chv

    Dividing both sides by chX to get the slope yields:

        chY   v (chu/chX) - u (chv/chX)
        --- = -------------------------
        chX           v² + v chv

        dY          v (chu/chX) - u (chv/chX)    v du/dX - u dv/dX
        -- = lim   (-------------------------) = -----------------
        dX   chX->0          v² + v chv                  v²

                    since chv-> 0 as chX-> 0. QED

Rule 7: Chain rule: Y = f(u(X)) => dy/dX = dY/du du/dX

    --Define e = chY/chu - dy/du, and note that e->0 as chu->0 since the slope and derivative converge to the same value.
    --Rearrange terms: chY = (dY/du)chu + e chu.
    --Divide by chX:

        chY   dY chu    chu
        --- = ------ + e---
        chX   du chX    chX

        dY          dY chu    chu    dY du    du   dY du
        -- = lim   (------ + e---) = -- -- + e-- = -- --
        dX   chX->0 du chX    chX    du dX    dX   du dX

                    since e-> 0 as chX-> 0. QED

Rule 2: Power rule: Y = Xⁿ => dY/dX = nX^n-1
        Note: This proof relies on several of the previous rules.
    This is proven for 3 cases: Case 1: n = a nonnegative integer
                                Case 2: n = a negative integer
                                Case 3: n = a rational number (a/b)

    Case 1: n = a nonnegative integer
    --Compute Y+chY = (X+chX)ⁿ as n gets bigger
        n=1: Y+chY = (X+chX)¹ = X + chX
        n=2: Y+chY = (X+chX)² = X² + 2X chX + chX²
        n=3: Y+chY = (X+chX)³ = (X+chX)²(X+chX)
                              = (X² + 2X chX + chX²)(X+chX)
                              = X³ + 3X²chX + 3X(chX)² + (chX)³
        Factor chX² out of
        the last 2 terms:     =   X³ + 3X²chX  + (3X + chX)(chX)²
 
        n>3 => Y+chY = (X+chX)n = Xⁿ + 3X^n-1chX + (terms in X + chX)(chX)² 
        (We just add more terms with
        chX raised to even higher powers)
    --Define T = (terms in X + chX)
    --Compute chY = (X+chX)ⁿ - Xⁿ
                  = Xⁿ + nX^n-1chX + T(chX)² - Xⁿ
                  =      nX^n-1chX + T(chX)²
    --Divide by chX to compute slope:
        chY   nX^n-1chX + T(chX)²
        --- = ----------------- = nX^n-1 + T(chX)
        chX         chX
    --let chX->0:

        dY         
        -- = lim    nX^n-1 + T(chX) = nX^n-1. QED
        dX   chX->0 

    Case 2: n = a negative integer
                            1
    --Rearrange terms: Y = ---, where -n > 0 since n < 0.
                           X^-n
    --Apply the quotient rule whre u=1 and v=X^-n:

        dY   X^-nd(1)/dX - 1 d(X^-n)/dX   0 - (-nX^-n-1)
        -- = ----------------------- = ------------ = nX^2n-n-1 = nX^n-1. QED
        dX          (X^-n)²                 X^-2n

    Case 3: n = a/b, a rational number

    --Raise both sies to the b power: Y = X^a/b => Y^b = X^(a/b)b = X^a
    --Rearrange terms to set up an implicit function: Y^b - X^a = 0
    --Use the chain rule to differentiate the implicit function:

        d(Y^b - X^a)/dX = bY^b-1 dY/dX - aX^a-1 dX/dX = bY^b-1 dY/dX - aX^a-1 = 0.

    --Rearrange trems to isolate dY/dX:

        bY^b-1 dY/dX - aX^a-1 = 0 => dY/dX = aX^a-1 / bY^b-1 = (a/b)X^a-1Y^1-b.

    --Substitute back the original expression for Y and rearrange terms:

        dY/dX = (a/b)X^a-1Y^1-b = (a/b)X^a-1(X^a/b)^1-b = (a/b)X^a-1X^a/b-a)
              = (a/b)X^a/b-1. QED