Higer order derivatives

Solo disponible en BuenasTareas
  • Páginas : 16 (3979 palabras )
  • Descarga(s) : 0
  • Publicado : 11 de diciembre de 2010
Leer documento completo
Vista previa del texto
mathematics joint project nº1 |
HIGHER-ORDER DERIVATIVES |
AND THEIR APPLICATION IN MATHEMATICS AND ECONOMICS |
|
HIGHER ORDER DERIVATIVES
We have to study the following function: y = x3-6x2+9x+4. And in order for that, we must study the first-order derivative when it is zero, that is to say, the possible critical points.
dydx = 3x2-12x+9
In mathematics, we can write in differentways the first-order derivative we are calculating, for instance, we can write the first-order derivative as f(x)'. Another way of writing the first-order derivative is dydx, as in the example. This way is more likely to appear in physics’ book. This way of writing, dydx, appears always in classical and quantum mechanics. It also appears as potential energy: -dfdt and it is represented by u, that isto say: u = -dfdt.
The first derivative in mathematics is equivalent to the marginal costs, income…in economics whereas in physics the first derivative refers to Speed/velocity and the second derivative means acceleration.
So the critical points in this function are:
f'(x)= 0
3x2-12x+9=0
x=12±144-108126 = 12±66 = 1 or 3
Thus, the points A (1, 8) and B (3, 4) are critical points. Afterknowing which the critical points are, we will study if they are either a local maxima or local minima. This will be done when we have already calculated the second-order derivative. There are no absolute maximums or minimums because it is a third-degree function.
The second-order derivative is:
f ''(x) = ddydxdx = 6x-12
We can write the second-order derivative as f ''(x), ddydxdx or as d2ydx2,this new function is also known as the second-order derivative of y with respect to x. You must remember that the second-order derivative tells us the gradient of f(x).
f ''(1) = 6(1)-12 = -6 < 0
Thus, the point A (1, 8) is a local maximum of the function f(x) = x3-6x2+9x+4. The value of y (8) at the point A is larger than at any other pointon the neighborhood.
f ''(3) = 6(3)-12= 6 > 0
Hence, the point B(3,4) is a local minimum of the function
f(x) = x3-6x2+9x+4.
The value of y (4) at the point B is smaller than at any other point on the neighborhood.
When x tends to infinite: lim x → ∞ f(x) → ∞
and when x tends to minus infinite: lim x → -∞ f(x) → -∞.
So the graph of fx= x3-6x2+9x+4 is the following:

The graph of thesecond-order derivative of y with respect to x is the following:

Stationary points
Second-order derivatives are used to find the critical points, also called stationary points, and in particular the turning-points of function curves. Stationary points can be either local maxima or local minima. A local maximum is a point which value of y is larger than any other point on its neighborhood. Onthe contrary, a local minimum is a point which value of y is smaller than any other point on its neighborhood.
We can see that for a local minimum or a local maximum, the tangent to the curve at one of these points must be horizontal. That is to say, the tangent of the curve must be zero, parallel to the x axis: f'(x) = 0. Local maxima and minima are known as turning-points.
When we are studyinga local maximum, we observe that on the left side of the critical point, the neighborhood’s points have a positive gradient and if we observe the other side the gradient of the neighborhood’s points is negative (the gradient at the local maximum is zero). So, when f'(x) is decreasing as x increases then f ''(x) must be negative.
Hence a local maximum has to carry out the following conditions:f'(x) = 0 and f ''(x) < 0

On the contrary, when we are studying a local minimum, we observe that on the left side of the critical point, the neighborhood’s points have a negative gradient and if we observe the other side the gradient of the neighborhood’s points is positive. So when f'(x) is increasing as x increases then f ''(x) must be positive.
Therefore a local minimum has to carry out...
tracking img