(c) A inction at, y) is dierentiable at (a, b) if and only if the linear approximation is a good approxi- mation of the function near the point (a, b). In other words, the surface 2 = f(l':, 3;) near (a, b, f[a, b)) looks like the tangent plane. In this part we will show that the existence of the partial derivatives is not enough to guarantee di'erentiability In part (b) you showed the partial derivative exists, however it turns out that f is not differentiable at [0, 0) as we will argue here. We will show the linear approximation is not a good approximation to z = f(a;,y) near (0, 0). To see why, write down the linear approximation for z = f[:r, y) using the partial derivatives you found in part (b). If f is di'erentiable then this linear approximation would be a good approximation of z = f[!l':, y) near (0, U), which means 2 = x, 3;) would look very much like this plane if you zoom in close enough at (0, 0). However, is z = f[:r, y) approximated by this plane along y = :1: near (3:, y) = (0, 0)? (Hint: Consider approaching [0,1]} along the line y = :3. How does f behave along this line? Is f approximated by the tangent plane along this line? (d) The Theorem in our notes (Section 14.4) says that z = f [:r, y) is di'erentiable at (a, b) if ft and fy are continuous at (a, b). For the example here, show directly that JI"m and 3\Consider the function 2xy(cty) if (x, y) # (0, 0) f(z, y) = 10 if (x, y) = (0,0) (a) Show that f is continuous at (0, 0). (Hint: you'll need to show lim(r,y)- (0,0) f (x, y) = f(0, 0).) (b) Show that fx (0, 0) = 0 by using the definition of the x-partial derivative: fr(0, 0) = limp _f(0th,0)-f(0,0) h Similarly, show fy(0, 0) = 0 by using the definition of the y-partial derivative