PLEASE SHOW STEPS. Consider: B) Prove that f is not differentiable at (0, 0), th
ID: 2866828 • Letter: P
Question
PLEASE SHOW STEPS.
Consider:
B) Prove that f is not differentiable at (0, 0), that is,
where L(x,y) is the linearization of f at (0,0).
You should use a computer to graph both f(x, y) and L(x, y) together, and zoom in closer and closer to the origin, to see what the problem is.
PLEASE SHOW STEPS. Consider: A) Let v= be a unit vector.we can write a = cos ?, b = sin ?. Compute Dvf(0,0) Use the definition of a directional derivative. This shows that Dvf(0,0) exists for every unit direction v. B) Prove that f is not differentiable at (0, 0), that is, where L(x,y) is the linearization of f at (0,0). You should use a computer to graph both f(x, y) and L(x, y) together, and zoom in closer and closer to the origin, to see what the problem is.Explanation / Answer
We prove that for an arbitrary point (x0; y0) we
have: D~uf (x0; y0) = fx (x0; y0) a+fy (x0; y0) b. Let us dene the function
g (h) = f (x0 + ha; y0 + hb) Then, we see that
g0 (0) = lim
h->
g (h) - g (0)/h
lim
h->0
f (x0 + ha; y0 + hb) - f (x0; y0)/
h
= D~uf (x0; y0)
On the other hand, we can also write g (h) = f (x; y) where x = x0 + ah
and y = y0 + bh. f is a function of x and y. But since both x and y
are functions of h, f is also a function of h. Using the chain rule (see
previous section), we have
g0 (h) = df/dh
Since this is true for any (x0; y0), the result follows.
so it will true for Since !u is a unit vector, if it makes an angle
with the positive x-axis, we can write !u = (cosx ; sinx ). The result follows