Multivariable Calculus: Finding g'(0)

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SUMMARY

The discussion focuses on the application of implicit differentiation in multivariable calculus, specifically in finding g'(0) and g''(0) using the function F(x, y). The user queries the rationale behind substituting x=0 and y=0 when calculating Fx(0,0) and Fy(0,0). It is established that since g(0)=0, substituting x=0 directly leads to y=g(0), confirming that y must also equal zero at that point. The Chain Rule is utilized to derive the relationship dy/dx = -Fx/Fy.

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  • Understanding of implicit differentiation
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  • Knowledge of multivariable functions
  • Basic concepts of derivatives in calculus
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Homework Statement


Here is the question with the solution:
http://dl.dropbox.com/u/64325990/MATH%20253/midterm%202.PNG

I don't understand how to do a. How did they know to use x=0 and y=0 for Fx(0,0) and Fy(0,0)?

I understand that this is implicit differentiation using Chain Rule where
dy/dx = -Fx/Fy

But g'(0) only tells us that x=0 but tells us nothing about y. So how would they know to use y=0 as well?Edit: Also does anyone have an idea how to start part b g''(0)?
 
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[STRIKE]Plug x=0 into F(x,g(x))=2 and solve for y=g(x).[/STRIKE]
EDIT: Actually, you can tell just from the fact that we let y=g(x) for the funtion F(x,y), and we are told that g(0)=0. y=g(x) is dependent on x. And when x=0, y=g(0), which we are told is equal to zero.
 
Last edited:

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