Expert: Paul Klarreich - 9/7/2007

Question
Find the slope of the tangent to the curve
y = 3 + 4x^2 - 2x^3 at the point where x = a.

Answer
Questioner:   Christine
Category:  Calculus
Private:  No
 
Subject:  derivatives and rates of change (long-hand version)
Question:  Find the slope of the tangent to the curve
y = 3 + 4x^2 - 2x^3 at the point where x = a.
..............................................
Hi, Christine,

I am not sure what you mean by the 'long-hand version.'  I think you mean that just writing:

dy/dx = 8x - 6x^2  and at  x = a,
dy/dx = 8a - 6a^2, which is the slope.

Is NOT what you want.  You mean to use the
A. Definition of the derivative.
A. Limit process.
A. Delta-process.
(names for the same thing.)

which means:

WARNING: USE COURIER FONT TO VIEW THIS.
     f(x) - f(a)
lim   ----------- =
x->a     x - a

     3 + 4x^2 - 2x^3 - (3 + 4a^2 - 2a^3)
lim   ----------------------------------- =
x->a               x - a

     3 + 4x^2 - 2x^3 - 3 - 4a^2 + 2a^3
lim   ---------------------------------- =
x->a                x - a

      4x^2 - 2x^3 - 4a^2 + 2a^3
lim   --------------------------- =
x->a            x - a

      4x^2 - 4a^2 - 2x^3 + 2a^3
lim   --------------------------- =
x->a             x - a

      4(x^2 - a^2) - 2(x^3 - a^3)
lim   ---------------------------- =
x->a             x - a

      4(x - a)(x + a) - 2(x - a)(x^2 + ax + a^2)
lim   ------------------------------------------- =
x->a             x - a

      
lim    4(x + a) - 2(x^2 + ax + a^2) =
x->a         

4(a + a) - 2(a^2 + aa + a^2) =

4(2a) - 2(3a^2) = 8a - 6a^2