Show that f(x,y) is differentiable

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SUMMARY

This discussion focuses on demonstrating the differentiability of the function f(x,y) through a specific equation involving Δx and Δy. The equation to prove is (1+Δx) + (1+Δx)sin(pi+Δy) - 1 = Δx - Δy + ε(Δx,Δy)Δx + ε(Δx,Δy)Δy. Key transformations include recognizing that sin(pi+Δy) simplifies to -sin(Δy), leading to the expression Δx - sin(Δy) - Δxsin(Δy). The discussion emphasizes the need to define ε(Δx,Δy) functions that converge to zero as (Δx,Δy) approaches (0,0).

PREREQUISITES
  • Understanding of multivariable calculus, specifically differentiability.
  • Familiarity with Taylor series expansions and limits.
  • Knowledge of trigonometric identities, particularly sin(π + y).
  • Proficiency in mathematical notation and epsilon-delta definitions.
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  • Study the concept of differentiability in multivariable functions.
  • Learn about the epsilon-delta definition of limits in calculus.
  • Explore Taylor series expansions for functions of multiple variables.
  • Investigate trigonometric identities and their applications in calculus.
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Students and educators in mathematics, particularly those studying calculus and analysis, as well as anyone involved in proving properties of multivariable functions.

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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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 $(\Delta x,\Delta y) \to (0,0).$)
 
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?
 

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