An equation is defined as x+y+x^5−y^5=0. a. Show that the equation determines y as a function of x in a neighbourhood of the origin (0,0). b. Denote the function from (a) by y=varphi(x). Find varphi^(5)(0) and varphi^(2004)(0). For the first part,I simply needed to prove that the equation is C^1 in the neighbourhood of the origin and that the first derivative w.r.t.y does not vanish at that point. These are the sufficient conditions for the existence of an implicit function in the neighbourhood of the origin. For the second part, I could do implicit differentiation (in a usual way) had it been for lower orders. But I need to learn the case for higher orders like the question asks. Please give me some hints.

spatularificw2 2022-09-28 Answered
An equation is defined as x + y + x 5 y 5 = 0.
a. Show that the equation determines y as a function of x in a neighbourhood of the origin (0,0).
b. Denote the function from ( a ) by y = φ ( x ). Find φ ( 5 ) ( 0 ) and φ ( 2004 ) ( 0 ).
For the first part,I simply needed to prove that the equation is C 1 in the neighbourhood of the origin and that the first derivative w.r.t.y does not vanish at that point. These are the sufficient conditions for the existence of an implicit function in the neighbourhood of the origin.
For the second part, I could do implicit differentiation (in a usual way) had it been for lower orders. But I need to learn the case for higher orders like the question asks. Please give me some hints.
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Answers (1)

Terahertztl
Answered 2022-09-29 Author has 8 answers
After one derivative, you have
1 + y + 5 x 4 5 y 4 y = 0.
Now, to find higher derivatives, make sure that you think about the chain rule. The implicit second derivative equation is
y + 20 x 3 20 y 3 ( y ) 2 5 y 4 y = 0.
The third derivative equation:
y + 60 x 2 60 y 2 ( y ) 3 40 y 3 y y 20 y 3 y y 5 y 4 y = 0.
Notice how several of the terms have a polynomial factor. With each derivative, the degree decreases. Can you see what happens after 5 derivatives? In the long run?
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