Checking differentiability \302\251 2006, Mike May, S.J. - maymk@slu.edurestart;We say a function of two variables is differentiable at a point if the graph near that point can be approximated by the tangent plane. A harder question is how to tell when a function given by a formula is differentiable. This worksheet examines how to check if a function is differentiable at a point.
<Text-field style="Heading 1" layout="Heading 1">Taking care of the easy points - nice function</Text-field>A standard theorem states that a function is differentiable at a point if both partial derivatives are defined and continuous at that point. Recall that polynomials are continuous functions. Similarly we get a continuous function if we take the sum, difference, product, or composition of two continuous functions. We can also take quotients if the denominator is not 0 and Maple will compute roots if the function is not negative. We also recall that the derivative is not defined when we take a root of 0 or (for even roots) a negative number. Thus with a nice function that is defined in a single algebraic case, we easily note that the function is differentiable at all points, except perhaps at the points where we divide by 0 or take a root of 0 or a negative number.
<Text-field style="Heading 1" layout="Heading 1">Exercise:</Text-field>1) The following functions are from the exercises in the section on differentiability in McCallum's Multivariable Calculus book. For each function, the reasoning above lets us say the function is "nice" everywhere, except perhaps at a small set of points. For each function identify the "difficult" points.(a) f(x, y) = x/y + y/x if x*y \342\211\240 0 and f(x, y) = 0 if x*y = 0.(b) f(x, y) = 2*x*y/(x^2 + y^2)^2 if LUkjbWlHNiMvSSttb2R1bGVuYW1lRzYiSSxUeXBlc2V0dGluZ0dJKF9zeXNsaWJHRic2JVEkYD9gRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0Yn and f(0, 0) = 0.(c) f(x, y) = x*y/sqrt(x^2 + y^2) if 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 and f(0, 0) = 0.(d) f(x, y) = x^2*y/(x^4 + y^2) if 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 and f(0, 0) = 0.(e) f(x, y) = x*y^2/(x^2 + y^2) if 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 and f(0, 0) = 0.(f) f(x, y) = x*y^2/(x^2 + y^4) if 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 and f(0, 0) = 0.(g) f(x, y) = sqrt(abs(x*y)).(h) f(x, y) = x*y*(x^2 - y^2)/(x^2 + y^2) if 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 and f(0, 0) = 0.
<Text-field style="Heading 1" layout="Heading 1">Really bad points - Where the function is not continuous.</Text-field>A function can only be differentiable where it is continuous. You may need to review the section on continuity. Recall that a function is continuous at a point if the limit at that point equals the value of the function at that point. Since we are only worried about difficult points, the function will often be defined by a special rule at the point, and by a formula nearby. Consider the function defined in example 4 , 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, unless (LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYnLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIixGJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRjEvJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdRLDAuMzMzMzMzM2VtRictRiw2JVEieUYnRi9GMi8lK2JhY2tncm91bmRHUS5bMjU1LDI1NSwyNTVdRidGOQ==) is the origin, in which case we use the special rule 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. The problem is at the origin. The first way to check continuity is to look at a graph to see if the function jumps.plot3d(x*y/(x^2+y^2),x=-1..1, y=-1..1,view=-1..1,
axes=boxed, style=patchcontour, shading=z);It is clear from looking at the picture that this function is not continuous at the origin. If we look down the z-axis it is clear that all the different contours extend in to the origin.(Technical tip - When looking at functions that might be discontinuous, it is often good to specify the view, or z-range. Otherwise the graph of a function going to infinity can eliminate the other details of the graph. It is also useful to use the shading=z option the that the color scheme is determined by z-value. At discontinuous points lots of colors try to come together.) We would also like an algebraic way to look at continuity. To do that we pick two paths to the trouble spot that seem to follow different contours. We can then take old fashioned limits on the paths and see that they have different limits. (We may find a path that has no limit at all.) For our example we pick the lines y=x and y=-x to approach the origin along. Maple lets us evaluate the limits either symbolically or graphically. func:= (x, y) -> x*y/(x^2+y^2);
path1 := func(x,-x);
path2 := func(x,x);
LimitOnPath1 := limit(path1, x=0);
LimitOnPath2 := limit(path2, x=0);
plot({path1,path2},x=-1..1, y=-1..1);Since the limits as we approach the origin of the two paths are different, the function cannot be continuous at the origin.
<Text-field style="Heading 1" layout="Heading 1">Exercise:</Text-field>2) Use a computer plot to show that the function defined by f(x, y) = x^2*y/(x^4 + y^2) when (x, y) \342\211\240 (0, 0) and f(0, 0) = 0is not continuous at the origin. Find two paths to the origin so that the limits on the paths are different from each other.LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=
<Text-field style="Heading 1" layout="Heading 1">Difficult points that are differentiable - continuous partial derivatives</Text-field>Having dealt with the nice points and the really bad points, we are ready to start working in from the ends. Typically, we are looking at points where the main formula would have division by 0, so the function is defined by a second rule at those points. If the function is continuous at that point, the correct value was chosen to bridge the gap. We will use the function defined by f(x, y) = x*y*(x^2 - y^2)/(x^2 + y^2) away from the origin and f(0, 0) = 0. The origin is the difficult point in this problem. A look at the graph shows that we have patched the hole at the origin with the correct value so the function is continuous.func := x*y*(x^2-y^2)/(x^2+y^2);
funcx := simplify(diff(func,x));
funcy := simplify(diff(func,y));Since we want to determine if the partial derivatives are continuous, the obvious approach is to look at the graphs of the partial derivatives. plot3d(funcx, x=-1..1, y=-1..1,view=-1..1, axes=BOXED);
plot3d(funcy, x=-1..1, y=-1..1,view=-1..1, axes=BOXED);Looking at the graphs it is clear that the partial derivatives are continuous at the origin. By the theorem from the book, the function is differentiable at the origin, hence everywhere.
<Text-field style="Heading 1" layout="Heading 1">Exercise:</Text-field>3) Use the technique described above to show the function defined by 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 if 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 and 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 is differentiable everywhere.
<Text-field style="Heading 1" layout="Heading 1">Continuous but not differentiable for lack of partial derivatives</Text-field>As we start working on functions that are continuous but not differentiable, the easiest ones are those where the partial derivatives are not defined. These are function that are not differentiable when we take a cross section in x or y The easiest examples involve absolute values and roots.Let 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. The function is continuous everywhere. When we take the cross sections x=0 and y=0 we get LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkobWZlbmNlZEdGJDYoLUkjbWlHRiQ2JVEieUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GNlEnbm9ybWFsRicvSSttc2VtYW50aWNzR0YkUSRhYnNGJy8lJW9wZW5HUSkmdmVyYmFyO0YnLyUmY2xvc2VHRj9GOi8lK2JhY2tncm91bmRHUS5bMjU1LDI1NSwyNTVdRidGOA== and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkobWZlbmNlZEdGJDYoLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GNlEnbm9ybWFsRicvSSttc2VtYW50aWNzR0YkUSRhYnNGJy8lJW9wZW5HUSkmdmVyYmFyO0YnLyUmY2xvc2VHRj9GOi8lK2JhY2tncm91bmRHUS5bMjU1LDI1NSwyNTVdRidGOA== respectively. Both of these curves have corners, so they are not differentiable at 0. Hence the partial derivatives of f(x,y) don't exist. func := (x, y) -> abs(x + y);
pathx := func(h,0);
pathy := func(0,h);
PartialOnPathx := diff(pathx, h);
LimitOnPathx := limit(PartialOnPathx, h=0);
PartialOnPathy := diff(pathy, h);
LimitOnPathy := limit(PartialOnPathy, h=0);
plot({PartialOnPathx,PartialOnPathy,pathx,pathy},h=-1..1, y=-1..1);
<Text-field style="Heading 1" layout="Heading 1">Exercise:</Text-field>4) Show that the function 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 is not differentiable at the origin.
<Text-field style="Heading 1" layout="Heading 1">The hard case - showing non-differentiability for a continuous function.</Text-field>The text points out that a function can be differentiable even if the partial derivatives are not continuous. We want some way to show that a function is not differentiable.The trick is to notice that for a differentiable function, all the tangent vectors at a point lie in a plane. Suppose that F(x,y) is differentiable at (a,b) with 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 and 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.Let U=[u_1, u_2] be a unit vector. Then the directional derivative in direction U must be the same whether we compute it from the definition of the directional derivative or by taking the dot product of U with the gradient vector.Consider the function defined by f(x,y) = (x^2*y)/(x^2+y^2).The comments about easy points tell us that this function is differentiable everywhere except perhaps at the origin. Look at the graph of the function and its partial derivatives.func:= (x, y) -> x^2*y/(x^2+y^2);
funcx := simplify(diff(func(x,y),x));
funcy := simplify(diff(func(x,y),y));
plot3d(func(x,y), x=-1..1, y=-1..1,view=-1..1, axes=BOXED);
plot3d(funcx, x=-1..1, y=-1..1,view=-1..3, axes=BOXED);
plot3d(funcy, x=-1..1, y=-1..1,view=-3..1, axes=BOXED);