Show f is differentiable but partial derivatives are not continuous

[theneedtoknow]

Homework Statement

Define f: Rn --------> R as
f(x) = (||x||^2)*sin (1/||x||) for ||x|| ≠ 0
f(x) = 0 for ||x|| = 0
Show that f is differentiable everywhere but that the partial derivatives are not continuous.

Homework Equations

The Attempt at a Solution

Showing that it is differentiable everywhere
if it is differentiable at the origin then

lim h---> 0 of ( f(0+h) - f(0) - c•h )/ ||h|| = 0


f(o+h) = ||h||^2sin(1/||h||)
f(0) = 0
c = gradient of f at 0

breaking it up into individual limits we have
lim h--->o of f(0+h)/||h|| + lim h--->0 of f(o)/||h|| + lim h--->0 of c•h/||h||

f(o+h)/||h|| = ||h||sin(1/||h||) which goes to zero since the absolute value of sin(1/||h||) is bounded by 1
f(0) / ||h|| goes to zero since f(0) = 0

but what does c•h/||h|| go to as h ---> 0
my intuition tells me it goes to c...but if the function is diffable at 0 then it has to go to zero as well...so I'm not sure how to show its differentiable everywhere

as for showing that the partial derivatives are not continuous

to calculate the jth partial derivatives (the derivative of f with respect to xj) i rewrite the function as (∑(xn^2))sin (1/root[∑(xn^2)])

df / dxj (using the product rule) = 2xj * sin (1/root[∑(xn^2)]) + cos(1/root[∑(xn^2)])* -0.5(∑(xn^2))^-1.5 * 2xj * [∑(xn^2)]

now...i can see that the particle derivatives are not continuous at ||x|| = 0 since the term (∑(xn^2))^-1.5 would be undefined if x1 to xn are all equal to zero

but am i also supposed to show that the limit as ||x|| approaches zero from the 2 sides of the particle derivatives is not equal? or what?

 

[mighty2000]

Thank you for posting your question. I am happy to help you with your inquiry about the differentiability and continuity of the function f: Rn --------> R.

First, let's start by proving that the function is differentiable everywhere. As you correctly stated, in order for a function to be differentiable at a point, the limit of the difference quotient must exist and equal to zero. In our case, the point in question is the origin (0,0) since the function is defined differently for ||x|| = 0 and ||x|| ≠ 0.

To prove differentiability at the origin, we need to show that the limit as h approaches 0 of the difference quotient (f(0+h) - f(0) - c•h)/||h|| equals zero. As you have already shown, the first two terms of the difference quotient go to zero as h approaches 0. Now, let's focus on the third term, c•h/||h||. Here, c represents the gradient of f at the origin, which we can calculate as the partial derivatives of f with respect to each variable at the origin. This means that c = (∂f/∂x1, ∂f/∂x2, ..., ∂f/∂xn) evaluated at the origin.

Now, let's look at the limit of c•h/||h|| as h approaches 0. Since c is a constant vector, we can pull it out of the limit and we are left with the limit of h/||h||. This limit is equal to 1, as ||h|| is just the magnitude of h, and h is a vector with all components approaching 0 as h approaches 0. Therefore, the limit of c•h/||h|| as h approaches 0 is equal to c.

In conclusion, the limit of the difference quotient as h approaches 0 is equal to c, which is the gradient of f at the origin. Since c is a constant vector, this limit exists and equals to zero, proving that the function f is differentiable at the origin. This same reasoning can be applied to any other point in Rn, showing that f is differentiable everywhere.

Now, let's move on to proving that the partial derivatives of f are not continuous. As you have correctly stated, the partial derivatives are not continuous at the origin, since the term