Mixing
difference between partial derivative at a point and composition of function
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I am doing some multivariable calculus and my mind is bugging on something :
We have a function : $f:mathbb{R}^2 to mathbb{R}$ which is $C^1$. Then we can calculate the partial derivative of $f$ with respect to $x$. This gives us a function : $mathbb{R}^2 to mathbb{R} :(x,y) mapsto frac{partial f}{partial x} (x,y)$.
From now one this is ok. But the problem is that there is a difference between : $$frac{partial f (rcos theta, r sin theta)}{partial x}$$ and $$frac{partial f }{partial x}(rcos theta, r sin theta)$$
right ?
To me the first one can be calculated using the chain rule. Since we taking the derivative with respect to $x$ of the function : $(r, theta) mapsto f circ (rcos theta, r sin theta)$, hence we are taking the derivative of the composition of two functions. Hence, if I am not mistaen we should have :
$$frac{partial f (rcos theta, r sin theta)}{partial x} = cos theta frac{partial f}{partial x} + sin theta frac{partial f}{partial x}$$
Yet the other expression is the derivative of $f$ with respect to $x$ evaluated at the point $(rcos theta, r sin theta)$. But now how can calculate this explicitly ?
For example if $f$ is such that : $forall t, f(tx) = tf(x)$ and I want to show that the partial derivative with respect to $x$ has the same property how can I do ?
I need to show that : $frac{partial f}{partial x} (tx) = t frac{partial f}{partial x} $ but the problem is that as I said before we don't have :
$$frac{partial f (tx)}{partial x} = frac{partial f}{partial x} (tx)$$
Hence I am bit lost and I don't understand what is really going between evaluating the derivtaive at a point and calculating the derivative of the tranpose.
thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
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add a comment |
$begingroup$
I am doing some multivariable calculus and my mind is bugging on something :
We have a function : $f:mathbb{R}^2 to mathbb{R}$ which is $C^1$. Then we can calculate the partial derivative of $f$ with respect to $x$. This gives us a function : $mathbb{R}^2 to mathbb{R} :(x,y) mapsto frac{partial f}{partial x} (x,y)$.
From now one this is ok. But the problem is that there is a difference between : $$frac{partial f (rcos theta, r sin theta)}{partial x}$$ and $$frac{partial f }{partial x}(rcos theta, r sin theta)$$
right ?
To me the first one can be calculated using the chain rule. Since we taking the derivative with respect to $x$ of the function : $(r, theta) mapsto f circ (rcos theta, r sin theta)$, hence we are taking the derivative of the composition of two functions. Hence, if I am not mistaen we should have :
$$frac{partial f (rcos theta, r sin theta)}{partial x} = cos theta frac{partial f}{partial x} + sin theta frac{partial f}{partial x}$$
Yet the other expression is the derivative of $f$ with respect to $x$ evaluated at the point $(rcos theta, r sin theta)$. But now how can calculate this explicitly ?
For example if $f$ is such that : $forall t, f(tx) = tf(x)$ and I want to show that the partial derivative with respect to $x$ has the same property how can I do ?
I need to show that : $frac{partial f}{partial x} (tx) = t frac{partial f}{partial x} $ but the problem is that as I said before we don't have :
$$frac{partial f (tx)}{partial x} = frac{partial f}{partial x} (tx)$$
Hence I am bit lost and I don't understand what is really going between evaluating the derivtaive at a point and calculating the derivative of the tranpose.
thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
I am doing some multivariable calculus and my mind is bugging on something :
We have a function : $f:mathbb{R}^2 to mathbb{R}$ which is $C^1$. Then we can calculate the partial derivative of $f$ with respect to $x$. This gives us a function : $mathbb{R}^2 to mathbb{R} :(x,y) mapsto frac{partial f}{partial x} (x,y)$.
From now one this is ok. But the problem is that there is a difference between : $$frac{partial f (rcos theta, r sin theta)}{partial x}$$ and $$frac{partial f }{partial x}(rcos theta, r sin theta)$$
right ?
To me the first one can be calculated using the chain rule. Since we taking the derivative with respect to $x$ of the function : $(r, theta) mapsto f circ (rcos theta, r sin theta)$, hence we are taking the derivative of the composition of two functions. Hence, if I am not mistaen we should have :
$$frac{partial f (rcos theta, r sin theta)}{partial x} = cos theta frac{partial f}{partial x} + sin theta frac{partial f}{partial x}$$
Yet the other expression is the derivative of $f$ with respect to $x$ evaluated at the point $(rcos theta, r sin theta)$. But now how can calculate this explicitly ?
For example if $f$ is such that : $forall t, f(tx) = tf(x)$ and I want to show that the partial derivative with respect to $x$ has the same property how can I do ?
I need to show that : $frac{partial f}{partial x} (tx) = t frac{partial f}{partial x} $ but the problem is that as I said before we don't have :
$$frac{partial f (tx)}{partial x} = frac{partial f}{partial x} (tx)$$
Hence I am bit lost and I don't understand what is really going between evaluating the derivtaive at a point and calculating the derivative of the tranpose.
thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
I am doing some multivariable calculus and my mind is bugging on something :
We have a function : $f:mathbb{R}^2 to mathbb{R}$ which is $C^1$. Then we can calculate the partial derivative of $f$ with respect to $x$. This gives us a function : $mathbb{R}^2 to mathbb{R} :(x,y) mapsto frac{partial f}{partial x} (x,y)$.
From now one this is ok. But the problem is that there is a difference between : $$frac{partial f (rcos theta, r sin theta)}{partial x}$$ and $$frac{partial f }{partial x}(rcos theta, r sin theta)$$
right ?
To me the first one can be calculated using the chain rule. Since we taking the derivative with respect to $x$ of the function : $(r, theta) mapsto f circ (rcos theta, r sin theta)$, hence we are taking the derivative of the composition of two functions. Hence, if I am not mistaen we should have :
$$frac{partial f (rcos theta, r sin theta)}{partial x} = cos theta frac{partial f}{partial x} + sin theta frac{partial f}{partial x}$$
Yet the other expression is the derivative of $f$ with respect to $x$ evaluated at the point $(rcos theta, r sin theta)$. But now how can calculate this explicitly ?
For example if $f$ is such that : $forall t, f(tx) = tf(x)$ and I want to show that the partial derivative with respect to $x$ has the same property how can I do ?
I need to show that : $frac{partial f}{partial x} (tx) = t frac{partial f}{partial x} $ but the problem is that as I said before we don't have :
$$frac{partial f (tx)}{partial x} = frac{partial f}{partial x} (tx)$$
Hence I am bit lost and I don't understand what is really going between evaluating the derivtaive at a point and calculating the derivative of the tranpose.
thank you !
real-analysis calculus multivariable-calculus partial-derivative
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
asked Jan 26 at 21:35
dghkgfzyukzdghkgfzyukz
16612
16612
add a comment |
add a comment |
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