MATH2111 Higher Several Variable Calculus (2 Viewers)

[InteGrand]

(Stats is done for this sem)

 

[leehuan]

Chain rule was what was on my mind, but I was getting lost at how to use it



I also started doubting myself, questioning if it's meant to be a Jacobian or just h-dash.


Unless, the answer is just that Jh(a) = 0 so RHS = 0 as well


Edit: Uh, nvm, found a gap in my learning.

 

[leehuan]

Found it really hard to work around the wording for this question. Not sure how to properly use the implicit function theorem.

Not too sure what exactly they want for my matrices (A|B) either

 

[seanieg89]

Edit: Uh, nvm, found a gap in my learning.

What was your end answer?

 

[RenegadeMx]

(Stats is done for this sem)

lel gl with calculus, most aids course

 

[leehuan]

What was your end answer?

I thought I had it but I probably didn't. Not sure what I could possibly do next after

I really don't feel like that should be the final answer either. Any further guidance?

lel gl with calculus, most aids course

dying

 

[seanieg89]

I thought I had it but I probably didn't. Not sure what I could possibly do next after

I really don't feel like that should be the final answer either. Any further guidance?

Idk what they want you to say with that wording and those assumptions really. Basically it just means that the image of f'(a) is a subspace of the kernel of g'(f(a)), or equivalently that the image of f'(a) is orthogonal to the gradient vector of g at f(a).

 

[leehuan]

Figures, should've thought in that direction
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[InteGrand]

Figures, should've thought in that direction
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What have you tried so far? Did you try maybe considering the inverse function theorem?

 

[leehuan]

What have you tried so far? Did you try maybe considering the inverse function theorem?

That was definitely the starting point, but that just asserted that there exists an open set that is open, not right?

Image of an open set under a continuous function might not be open

 

[InteGrand]

Image of an open set under a continuous function might not be open

Generally this is correct, but consider the implications of f being an invertible (locally) continuous map.

 

[InteGrand]

That was definitely the starting point, but that just asserted that there exists an open set that is open, not right?

Also, if V is a subset of U, then f(V) is automatically a subset of f(U) (true for any function and sets (that make sense)).

(Or maybe you meant that it doesn't state that f(V) is open.)

 

[leehuan]

Also, if V is a subset of U, then f(V) is automatically a subset of f(U) (true for any function and sets (that make sense)).

(Or maybe you meant that it doesn't state that f(V) is open.)

Yeah the latter lol
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Just a yes or no answer please because I can't just tell what to use immediately. I know that it converges pointwise to the zero function.



Because seeing as though it converges pointwise to something continuous I doubt I can use that, and the Weierstrass M-test doesn't even seem relevant.

 

[InteGrand]

Yeah the latter lol
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Just a yes or no answer please because I can't just tell what to use immediately. I know that it converges pointwise to the zero function.



Because seeing as though it converges pointwise to something continuous I doubt I can use that, and the Weierstrass M-test doesn't even seem relevant.

No

 

[leehuan]

 

[InteGrand]

Also the first term on your RHS has a mistake/typo (shouldn't have y).

 

[glittergal96]

chain rule.

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