Question

"Show that (x,y)=x2+3y is differentiable at every point. In

other words, show that Δz=(x+Δx,y+Δy)−(x,y)=xΔx+yΔy+ε1Δx+ε2Δ

Answer 1

To show that the function (x,y) = x² + 3y is differentiable at every point, we need to prove that the limit of the difference quotient exists as the independent variables approach 0. By simplifying and taking the limit, we find that the limit exists and is equal to 2x + 3(dy/dx), confirming the differentiability of the function.

Step-by-step explanation:

In order to show that a function is differentiable at every point, we need to show that the limit of the difference quotient exists as the independent variables approach 0. In this case, we need to show that the limit of the difference quotient Δz = xΔx + yΔy + ε1Δx + ε2Δy as Δx and Δy approach 0 exists. Let's proceed with the proof:

1. Start by substituting the given function into the difference quotient: Δz = (x + Δx)² + 3(y + Δy) - (x² + 3y).
2. Expand the terms and simplify: Δz = x² + 2xΔx + Δx² + 3y + 3Δy - x² - 3y.
3. Combine like terms: Δz = 2xΔx + Δx² + 3Δy.
4. Factor out Δx from the first two terms: Δz = Δx(2x + Δx) + 3Δy.
5. Now, divide both sides by Δx: Δz/Δx = 2x + Δx + 3(Δy/Δx).
6. Take the limit as Δx approaches 0: lim(Δz/Δx) = 2x + 3(dy/dx).

Since the limit exists and is equal to 2x + 3(dy/dx), we can conclude that the function (x,y) = x² + 3y is differentiable at every point.