Show that f(x,y) is differentiable

In summary, the problem is to show that (1+Δx) + (1+Δx)sin(pi+Δy) - 1 = Δx - Δy + ε(Δx,Δy)Δx + ε(Δx,Δy)Δy. By using the trigonometric identity sin(pi+Δy) = -sin(Δy), the left side can be simplified to Δx - sin(Δy) - Δxsin(Δy). To make it match the right side, we can write it as Δx - Δy + Δx(-sin(Δy) + Δy(1
  • #1
Jamie2
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Problem:
View attachment 2105

I plugged in fx, fy, and f(1,pi) everywhere I could but I have no idea how to move on from here. I'm stuck trying to show that:

(1+Δx) + (1+Δx)sin(pi+Δy) - 1 = Δx - Δy + ε(Δx,Δy)Δx + ε(Δx,Δy)Δy
 

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  • #2
Jamie said:
I'm stuck trying to show that:

(1+Δx) + (1+Δx)sin(pi+Δy) - 1 = Δx - Δy + ε(Δx,Δy)Δx + ε(Δx,Δy)Δy
You are nearly there! You want to show that $$(1+\Delta x) + (1+\Delta x)\sin(\pi+\Delta y) - 1 = \Delta x - \Delta y + \varepsilon_1(\Delta x,\Delta y)\Delta x + \varepsilon_2(\Delta x,\Delta x y)\Delta y.$$ On the left side of that equation, $ \sin(\pi+\Delta y) = - \sin(\Delta y)$. So the left side of the equation becomes $$\Delta x - \sin(\Delta y) - \Delta x\sin(\Delta y).$$ You want that to look like the right side of the equation. So write it as $$\Delta x - \Delta y + \Delta x(-\sin(\Delta y) + \Delta y\left(1 - \frac{\sin(\Delta y)}{\Delta y}\right).$$ Now can you see how to choose the functions $\varepsilon_1(\Delta x,\Delta y)$ and $\varepsilon_2(\Delta x,\Delta y)$? (Remember that you have to show that they go to $0$ as $(\Delta x,\Delta y) \to (0,0).$)
 
  • #3
Opalg said:
You are nearly there! You want to show that $$(1+\Delta x) + (1+\Delta x)\sin(\pi+\Delta y) - 1 = \Delta x - \Delta y + \varepsilon_1(\Delta x,\Delta y)\Delta x + \varepsilon_2(\Delta x,\Delta x y)\Delta y.$$ On the left side of that equation, $ \sin(\pi+\Delta y) = - \sin(\Delta y)$. So the left side of the equation becomes $$\Delta x - \sin(\Delta y) - \Delta x\sin(\Delta y).$$ You want that to look like the right side of the equation. So write it as $$\Delta x - \Delta y + \Delta x(-\sin(\Delta y) + \Delta y\left(1 - \frac{\sin(\Delta y)}{\Delta y}\right).$$ Now can you see how to choose the functions $\varepsilon_1(\Delta x,\Delta y)$ and $\varepsilon_2(\Delta x,\Delta y)$? (Remember that you have to show that they go to $0$ as $(\Delta x,\Delta y) \to (0,0).$)
thank you! I understand how to finish the problem now. But could you explain your simplification of the left side in a little more detail?
 

FAQ: Show that f(x,y) is differentiable

What does it mean for a function to be differentiable?

When a function is differentiable, it means that it has a well-defined derivative at every point in its domain. In other words, the function is smooth and has a tangent line at every point.

How do you prove that a function is differentiable?

To prove that a function is differentiable, you need to show that the limit of the difference quotient (the slope of a secant line) approaches a specific value as the distance between the two points on the curve approaches zero. This value is known as the derivative of the function at that point.

What is the relationship between continuity and differentiability?

A function must be continuous in order to be differentiable. This means that the function must be defined and have a limit at every point in its domain. A function can be continuous without being differentiable, but it cannot be differentiable without being continuous.

What are the key properties of a differentiable function?

A differentiable function is continuous, has a well-defined derivative at every point in its domain, and is smooth (has a tangent line) at every point. It also satisfies the product rule, chain rule, and sum rule.

How is differentiability related to the concept of local linearity?

A differentiable function is locally linear, meaning that the function can be approximated by a linear function (a straight line) at every point in its domain. The derivative of the function at that point is equal to the slope of the tangent line, which is the best linear approximation of the function at that point.

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