How to determine the $dy$-term of the differential of $f(x,y,z)= sin(x+y)sin(y+z)$ to be $sin(x+2y+z)$?
$begingroup$
So I want to know how I can simplify the $dy$-part of the differential of $f(x,y,z)= sin(x+y)sin(y+z)$.
I already have the $dx$-term which is $[cos(x+y)sin(y+z)]dx$ and the $dz$-term which becomes $[sin(x+y)sin(y+z)]dz$.
So I already know that the differential of the entire thing is
$df(x,y,z)= [cos(x+y)sin(y+z)]dx+ sin(x+2y+z)dy + [sin(x+y)cos(y+z)]dz$
The question actually is what is the $dfrac{partial f(x,y,z)}{partial y}$? I don't know any trigonemetric identity to of 3 terms into 1 sin. Can someone explain this to me?
EDIT: I already have for the $dy$-term that $cos(x+y)sin(y+z)+ cos(y+z)sin(x+y)]dy$
calculus derivatives partial-derivative
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
$endgroup$
add a comment |
$begingroup$
So I want to know how I can simplify the $dy$-part of the differential of $f(x,y,z)= sin(x+y)sin(y+z)$.
I already have the $dx$-term which is $[cos(x+y)sin(y+z)]dx$ and the $dz$-term which becomes $[sin(x+y)sin(y+z)]dz$.
So I already know that the differential of the entire thing is
$df(x,y,z)= [cos(x+y)sin(y+z)]dx+ sin(x+2y+z)dy + [sin(x+y)cos(y+z)]dz$
The question actually is what is the $dfrac{partial f(x,y,z)}{partial y}$? I don't know any trigonemetric identity to of 3 terms into 1 sin. Can someone explain this to me?
EDIT: I already have for the $dy$-term that $cos(x+y)sin(y+z)+ cos(y+z)sin(x+y)]dy$
calculus derivatives partial-derivative
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
$endgroup$
$begingroup$
If the question is $partial f / partial y$ why are you calculating the total differential ?
$endgroup$
– Rebellos
Dec 27 '18 at 9:47
$begingroup$
Because I need it for the total differential. Should I change the question to only the partial derivative?
$endgroup$
– Anonymous196
Dec 27 '18 at 9:48
add a comment |
$begingroup$
So I want to know how I can simplify the $dy$-part of the differential of $f(x,y,z)= sin(x+y)sin(y+z)$.
I already have the $dx$-term which is $[cos(x+y)sin(y+z)]dx$ and the $dz$-term which becomes $[sin(x+y)sin(y+z)]dz$.
So I already know that the differential of the entire thing is
$df(x,y,z)= [cos(x+y)sin(y+z)]dx+ sin(x+2y+z)dy + [sin(x+y)cos(y+z)]dz$
The question actually is what is the $dfrac{partial f(x,y,z)}{partial y}$? I don't know any trigonemetric identity to of 3 terms into 1 sin. Can someone explain this to me?
EDIT: I already have for the $dy$-term that $cos(x+y)sin(y+z)+ cos(y+z)sin(x+y)]dy$
calculus derivatives partial-derivative
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
$endgroup$
So I want to know how I can simplify the $dy$-part of the differential of $f(x,y,z)= sin(x+y)sin(y+z)$.
I already have the $dx$-term which is $[cos(x+y)sin(y+z)]dx$ and the $dz$-term which becomes $[sin(x+y)sin(y+z)]dz$.
So I already know that the differential of the entire thing is
$df(x,y,z)= [cos(x+y)sin(y+z)]dx+ sin(x+2y+z)dy + [sin(x+y)cos(y+z)]dz$
The question actually is what is the $dfrac{partial f(x,y,z)}{partial y}$? I don't know any trigonemetric identity to of 3 terms into 1 sin. Can someone explain this to me?
EDIT: I already have for the $dy$-term that $cos(x+y)sin(y+z)+ cos(y+z)sin(x+y)]dy$
calculus derivatives partial-derivative
calculus derivatives partial-derivative
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
edited Dec 27 '18 at 9:47
Anonymous196
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
asked Dec 27 '18 at 9:39
Anonymous196Anonymous196
8352926
8352926
$begingroup$
If the question is $partial f / partial y$ why are you calculating the total differential ?
$endgroup$
– Rebellos
Dec 27 '18 at 9:47
$begingroup$
Because I need it for the total differential. Should I change the question to only the partial derivative?
$endgroup$
– Anonymous196
Dec 27 '18 at 9:48
add a comment |
$begingroup$
If the question is $partial f / partial y$ why are you calculating the total differential ?
$endgroup$
– Rebellos
Dec 27 '18 at 9:47
$begingroup$
Because I need it for the total differential. Should I change the question to only the partial derivative?
$endgroup$
– Anonymous196
Dec 27 '18 at 9:48
$begingroup$
If the question is $partial f / partial y$ why are you calculating the total differential ?
$endgroup$
– Rebellos
Dec 27 '18 at 9:47
$begingroup$
If the question is $partial f / partial y$ why are you calculating the total differential ?
$endgroup$
– Rebellos
Dec 27 '18 at 9:47
$begingroup$
Because I need it for the total differential. Should I change the question to only the partial derivative?
$endgroup$
– Anonymous196
Dec 27 '18 at 9:48
$begingroup$
Because I need it for the total differential. Should I change the question to only the partial derivative?
$endgroup$
– Anonymous196
Dec 27 '18 at 9:48
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
1
$begingroup$
Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
$endgroup$
– Anonymous196
Dec 27 '18 at 9:50
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
1
$begingroup$
Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
$endgroup$
– Anonymous196
Dec 27 '18 at 9:50
add a comment |
$begingroup$
Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
1
$begingroup$
Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
$endgroup$
– Anonymous196
Dec 27 '18 at 9:50
add a comment |
$begingroup$
Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
$endgroup$
Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
answered Dec 27 '18 at 9:48
Thomas ShelbyThomas Shelby
4,7382727
4,7382727
1
$begingroup$
Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
$endgroup$
– Anonymous196
Dec 27 '18 at 9:50
add a comment |
1
$begingroup$
Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
$endgroup$
– Anonymous196
Dec 27 '18 at 9:50
1
1
$begingroup$
Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
$endgroup$
– Anonymous196
Dec 27 '18 at 9:50
$begingroup$
Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
$endgroup$
– Anonymous196
Dec 27 '18 at 9:50
add a comment |
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$begingroup$
If the question is $partial f / partial y$ why are you calculating the total differential ?
$endgroup$
– Rebellos
Dec 27 '18 at 9:47
$begingroup$
Because I need it for the total differential. Should I change the question to only the partial derivative?
$endgroup$
– Anonymous196
Dec 27 '18 at 9:48