# Limits and differentiability

1. Nov 24, 2005

### freya81

Am mainly stuck on parts c) and d) but thought i'd put in the other questions as an aid

2. a) Define what it means to describe a function f of two real variables as differentiable at (a, b)? Define (as limits) the partial derivatives df/dx and df/dy at
(a, b) and prove that if f is differentiable at (a, b) then both these partial derivatives exist.

b) Prove from the definition in a) that the function f defined by f(x,y) =xy(x+y) is differentiable at every point of it’s domain

c) If g(x, y) = xy prove that g is not differentiable at (0, b) for any non- zero value of b

d) Prove that the function g of part c) is differentiable at all points (a,b) for which a is not zero and at the origin (0,0)

2. Nov 24, 2005

### freya81

it this correct
2. a) f is differentiable at (a,b) iff there exists a linear mapping L such that lim((f(a+h,b+k)-f(a,b)-L(h,k))/|(h,k)|,(h,k)->(0,0))=0. ∂f/∂x(a,b)=lim((f(a+h,b)-f(a,b))/h,h->0), and ∂f/∂y(a,b)=lim((f(a,b+k)-f(a,b))/k,k->0).
does L=grad f; that is, L(h,k)=(∂f/∂x,∂f/∂y).(h,k)=h*∂f/∂x+k*∂f/∂y.

what do i do next?

3. Nov 24, 2005

### HallsofIvy

Staff Emeritus
L is the derivative of f in the direction of the vector <h,k> (i.e. from (a,b) to (a+h,b+k). In a given coordinate system, L is the gradient of f, evaluated at (a,b), dot the vector <h,k>. Note, however, that L, as a linear mapping, exists without a coordinate system. The partial derivatives, of course, depend upon the coordinate system. It is also true, by the way, that a function can have partial derivatives at a point and not be differentiable there.

For b, c, and d, about all I can say is "do it"! You can find the partial derivatives of each so you know what L is for each one. Put the functions and L into the formula in a and do the algebra.

Last edited: Nov 24, 2005