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
asked Nov 13 at 6:22
Minh
165
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up vote
0
down vote
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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 Nov 13 at 6:22
Minh
165
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 Nov 13 at 6:22
Minh
165
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 Nov 13 at 6:22
Minh
165
asked Nov 13 at 6:22
Minh
165
asked Nov 13 at 6:22
Minh
165
asked Nov 13 at 6:22
Minh
165
asked Nov 13 at 6:22
Minh
165
165
add a comment |
add a comment |
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