Partial derivatives of the dependent variable.
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I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
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add a comment |
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I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
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Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
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– Devashish Kaushik
Dec 29 '18 at 10:02
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I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
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– Icchyamoy
Dec 29 '18 at 10:05
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I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
add a comment |
$begingroup$
I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
$endgroup$
I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
edited Dec 29 '18 at 10:55
Atheesh Rathnaweera
31
31
asked Dec 29 '18 at 9:51
IcchyamoyIcchyamoy
11
11
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
add a comment |
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06
add a comment |
1 Answer
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If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
$endgroup$
add a comment |
$begingroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
$endgroup$
add a comment |
$begingroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
$endgroup$
If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.
$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$
The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.
answered Dec 29 '18 at 10:11
Devashish KaushikDevashish Kaushik
638220
638220
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$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02
$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05
$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06