Does the second order differentiability always imply the first order differentiability?
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If one considers the one-variable function, we have the definition
$$f''(x):=(f'(x))'$$
So the second order derivative must be first provided the existence of the first order derivative.
But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is
$$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$
So if these imply the existence of $displaystylefrac{partial f}{partial y}$?
multivariable-calculus
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If one considers the one-variable function, we have the definition
$$f''(x):=(f'(x))'$$
So the second order derivative must be first provided the existence of the first order derivative.
But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is
$$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$
So if these imply the existence of $displaystylefrac{partial f}{partial y}$?
multivariable-calculus
asked yesterday
Minh
164
New contributor
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If one considers the one-variable function, we have the definition
$$f''(x):=(f'(x))'$$
So the second order derivative must be first provided the existence of the first order derivative.
But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is
$$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$
So if these imply the existence of $displaystylefrac{partial f}{partial y}$?
multivariable-calculus
asked yesterday
Minh
164
New contributor
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If one considers the one-variable function, we have the definition
$$f''(x):=(f'(x))'$$
So the second order derivative must be first provided the existence of the first order derivative.
But if we consider the multivariable funtion, namely $f(x,y),$ and we also have the existence $displaystyle frac{partial^2f}{partial xpartial y},$ that by definition is
$$frac{partial^2f}{partial xpartial y}:=frac{partialleft(frac{partial f}{partial x}right)}{partial y}$$
So if these imply the existence of $displaystylefrac{partial f}{partial y}$?
multivariable-calculus
multivariable-calculus
asked yesterday
Minh
164
New contributor
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Minh
164
New contributor
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Minh
164
New contributor
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Minh
164
asked yesterday
Minh
164
164
New contributor
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Minh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Minh is a new contributor. Be nice, and check out our Code of Conduct.
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