Let f(x, y)={ (x y-1)/(x^2+y^2-2) if x^2+y^2≠ 2 (1)/(2) if x^2+y^2=2 . Show tha

Let f(x, y)={ (x y-1)/(x^2+y^2-2) if x^2+y^2≠ 2 ...

Key Concepts

Path Dependency in Limit Evaluation

Evaluating limits along different paths is crucial in multivariable calculus because the existence of partial derivatives does not ensure that the function behaves uniformly in every direction. A function is not differentiable at a point if there exists at least one path along which the ratio of the error to the distance does not tend to zero, even if the partial derivatives themselves exist.

Linear Approximation and the Remainder Term

The concept of linear approximation involves expressing the change in a function as a sum of a linear part, which is given by the partial derivatives, and a remainder term. The differentiability of the function is confirmed if this remainder term is of higher order (i.e., negligible compared to the linear term) as the input approaches the point in question.

Partial Derivatives

Partial derivatives measure the rate at which a function changes as one of the variables is varied while keeping the other variables fixed. They are defined as the limit of the difference quotient along the axis of the variable of interest. Even if these derivatives exist at a point, they only capture directional changes along the coordinate axes and may not reflect the overall behavior of the function in all directions.

Differentiability in Multiple Variables

Differentiability for functions of several variables is defined by the existence of a linear map (often thought of as the tangent plane) that approximates the function in the neighborhood of a point, with an error term that becomes negligible relative to the size of the change in the input. A function may have partial derivatives at a point and yet fail to be differentiable if the remainder term does not vanish in a way that is uniform with respect to all directions.

Transcript

00:01 For this problem, we are told to let f of xy be the piecewise defined function, xy minus 1 divided by x squared plus y squared minus 2, if x squared plus y squared does not equal to, and 1 over 2 if x squared plus y squared equals 2.

00:16 We are then asked to show that the partial derivatives, with respect to x and y, at the point 1 -1, each exist, but that f is not differentiable at the point 1 -1.

00:28 So to begin with finding those partial derivatives, we know that by definition, fx of 1 -1, should be equal to the limit as delta -x approaches 0 of f of 1 plus delta -x, 1 minus f -of -1 -1 -1, all divided by delta -x.

00:54 We know that f -1 -1 equals 1 over 2.

00:57 So we can write this then as the limit, as delta x approaches zero, of, now f of 1 plus delta x, 1 would be, let's see here.

01:09 I'll write it down as 1 plus delta x times 1 minus 1, divided by 1 plus delta x all squared, plus 1 squared, so plus 1, minus 2.

01:25 So we can just write that as minus 1.

01:29 And we have minus 1 over 2, and we divide everything through by delta x.

01:35 So this will then be equal to the limit as delta x approaches 0 of, one moment here.

01:45 So simplifying the numerator, we can write that down as negative delta x over 4 plus 2 delta x, and then we're still dividing by delta x or multiplying by one over delta x.

01:59 So we can see that the delta x is divided out.

02:01 We're left with negative 1 over 4 plus 2 delta x, or the limit of that, as delta x approaches 0.

02:13 So we clearly get a result of negative 1 over 4 as the partial derivative.

02:17 Then taking the partial derivative with respect to y, we'll get that this will be the limit as delta y approaches 0.

02:25 Of f of, or it would be f of 1, then 1 plus delta y, minus f of 1 1, which we already know, that's 1 half, all divided by delta y.

02:43 Now, substituting in 1 plus delta y in place of y, we'd get 1 plus delta y minus 1 divided by 1 plus delta y, all squared, plus 1 minus 2, so minus 1, minus 1 over 2, all over delta y.

03:03 And i'll note we're still taking a limit as delta y approaches 0 here.

03:09 Oops, let me fix that line that showed up there.

03:15 Now we can see that the overall form of what we are looking at here is identical to what we saw for delta, or for the partial derivative with respect to x.

03:27 So we can conclude that we'll arrive at the same point, particularly being that partial derivative equals negative 1 over 4...

Please give Ace some feedback