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# Differentiability, linearization.

Theorem 5.5   Let f be a function defined on an open domain .Suppose that f has first order partial derivatives at every point of and that these partial derivatives are continuous at the point (x0,y0). Denote .

Then:

 (5.2.1)

where and .

Definition 5.6   A function f is differentiable at the point (x0,y0) if the partial derivatives fx(x0,y0) and fy(x0,y0) exist and the equation  2.1 holds for f at (x0,y0).

The function f is differentiable on the open domain if it is differentiable at every point of .

Corollary 5.7   If the first order partial derivatives of f are continuous on the open domain , then f is differentiable on .

Theorem 5.8   If f is differentiable at the point (x0,y0), then f is continuous at the point (x0,y0).

Definition 5.9   The linearization of a function f at a point (x0,y0) is the function

Lf(x0,y0) (x,y)=f(x0,y0)+ fx(x0,y0) (x-x0)+fy(x0,y0)(y-y0)

Example 5.10   Let and (x0,y0)=(1,0). Then and . Thus, the linearization of f at (1,0) is:

We can use this linearization to compute an approximation of f(1.02, 0.15):

The true'' value is .

Question: Can we have an estimation of the error when using the linearization instead of the actual function?

The positive answer is described in the following result:

Proposition 5.11   Suppose that f has continuous partial derivatives of order 1 and 2 inan open domain . Let be a rectangle centerd at the point (x0,y0) and contained in . We denote by M an upper bound for |fxx|, |fxy| and |fyy|. Then the error made when replacing f(x,y) by its linearization for satisfies the following inequality:

Example 5.12   Take f(x,y)=x2y+xy3-1, (x0,y0)=(1,2) and .

We have: fx=2xy+y3, fy=x2+3xy2, fxx= 2y, fxy= 2x+y2 and fyy=6xy.

On the rectangle , we can take: M=. Then we have the inequality:

As we have f(1,2)=9, the error percentage is less than , i.e. we made an error of at most .

Next: Partial derivatives of higher Up: Partial derivatives; the differential Previous: First partial derivatives.
Noah Dana-Picard
2001-05-30