How to determine the $dy$-term of the differential of $f(x,y,z)= sin(x+y)sin(y+z)$ to be $sin(x+2y+z)$?












1














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 $cos(x+y)sin(y+z)+ cos(y+z)sin(x+y)]dy$










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  • If the question is $partial f / partial y$ why are you calculating the total differential ?
    – Rebellos
    Dec 27 '18 at 9:47












  • Because I need it for the total differential. Should I change the question to only the partial derivative?
    – Anonymous I
    Dec 27 '18 at 9:48


















1














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 $cos(x+y)sin(y+z)+ cos(y+z)sin(x+y)]dy$










share|cite|improve this question
























  • If the question is $partial f / partial y$ why are you calculating the total differential ?
    – Rebellos
    Dec 27 '18 at 9:47












  • Because I need it for the total differential. Should I change the question to only the partial derivative?
    – Anonymous I
    Dec 27 '18 at 9:48
















1












1








1







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 $cos(x+y)sin(y+z)+ cos(y+z)sin(x+y)]dy$










share|cite|improve this question















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






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edited Dec 27 '18 at 9:47

























asked Dec 27 '18 at 9:39









Anonymous I

8302826




8302826












  • If the question is $partial f / partial y$ why are you calculating the total differential ?
    – Rebellos
    Dec 27 '18 at 9:47












  • Because I need it for the total differential. Should I change the question to only the partial derivative?
    – Anonymous I
    Dec 27 '18 at 9:48




















  • If the question is $partial f / partial y$ why are you calculating the total differential ?
    – Rebellos
    Dec 27 '18 at 9:47












  • Because I need it for the total differential. Should I change the question to only the partial derivative?
    – Anonymous I
    Dec 27 '18 at 9:48


















If the question is $partial f / partial y$ why are you calculating the total differential ?
– Rebellos
Dec 27 '18 at 9:47






If the question is $partial f / partial y$ why are you calculating the total differential ?
– Rebellos
Dec 27 '18 at 9:47














Because I need it for the total differential. Should I change the question to only the partial derivative?
– Anonymous I
Dec 27 '18 at 9:48






Because I need it for the total differential. Should I change the question to only the partial derivative?
– Anonymous I
Dec 27 '18 at 9:48












1 Answer
1






active

oldest

votes


















4














Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$






share|cite|improve this answer

















  • 1




    Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
    – Anonymous I
    Dec 27 '18 at 9:50













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$






share|cite|improve this answer

















  • 1




    Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
    – Anonymous I
    Dec 27 '18 at 9:50


















4














Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$






share|cite|improve this answer

















  • 1




    Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
    – Anonymous I
    Dec 27 '18 at 9:50
















4












4








4






Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$






share|cite|improve this answer












Hint:
$$sin (a+b)=cos(a)sin(b)+ cos(b)sin(a)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 9:48









Thomas Shelby

1,516216




1,516216








  • 1




    Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
    – Anonymous I
    Dec 27 '18 at 9:50
















  • 1




    Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
    – Anonymous I
    Dec 27 '18 at 9:50










1




1




Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
– Anonymous I
Dec 27 '18 at 9:50






Oh, oké. I'm so stupid that I didn't even search for that. Thank you for the hint.
– Anonymous I
Dec 27 '18 at 9:50




















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