Value of multivariable function












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We have $$f(x,y,z)=xz+x^2z+sin(x+2y+z).$$ What is the the value of $df(1,-1,1)$.
I found the partial derivatives of f and than what? Is something like $$df(a,b,c,)=frac{df}{dx}(a,b,c)+frac{df}{dy}(a,b,c)+frac{df}{dz}(a,b,c)?$$
Thx guys.






partial-derivative







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  • $begingroup$
    Not quite. Your left side, "$df(a,b,c)$" is a "differential" but the right side "$frac{partial f}{partial x}(a, b, c)+ frac{partial f}{partial y}(a, b, c)+ frac{partial f}{partial z}(a, b, c)$" is not. You want "$df(a,b,c)= frac{partial f}{partial x}(a, b, c)dx+ frac{partial f}{partial y}(a, b, c)dy+ frac{partial f}{partial z}(a, b, c)dz$".
    $endgroup$
    – user247327
    Dec 8 '18 at 11:47










  • $begingroup$
    Ok, I get it. Stil don t know the answer and how to find it.
    $endgroup$
    – Numbers
    Dec 8 '18 at 13:30
















0












$begingroup$


We have $$f(x,y,z)=xz+x^2z+sin(x+2y+z).$$ What is the the value of $df(1,-1,1)$.
I found the partial derivatives of f and than what? Is something like $$df(a,b,c,)=frac{df}{dx}(a,b,c)+frac{df}{dy}(a,b,c)+frac{df}{dz}(a,b,c)?$$
Thx guys.






partial-derivative







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not quite. Your left side, "$df(a,b,c)$" is a "differential" but the right side "$frac{partial f}{partial x}(a, b, c)+ frac{partial f}{partial y}(a, b, c)+ frac{partial f}{partial z}(a, b, c)$" is not. You want "$df(a,b,c)= frac{partial f}{partial x}(a, b, c)dx+ frac{partial f}{partial y}(a, b, c)dy+ frac{partial f}{partial z}(a, b, c)dz$".
    $endgroup$
    – user247327
    Dec 8 '18 at 11:47










  • $begingroup$
    Ok, I get it. Stil don t know the answer and how to find it.
    $endgroup$
    – Numbers
    Dec 8 '18 at 13:30














0












0








0





$begingroup$


We have $$f(x,y,z)=xz+x^2z+sin(x+2y+z).$$ What is the the value of $df(1,-1,1)$.
I found the partial derivatives of f and than what? Is something like $$df(a,b,c,)=frac{df}{dx}(a,b,c)+frac{df}{dy}(a,b,c)+frac{df}{dz}(a,b,c)?$$
Thx guys.






partial-derivative







share|cite|improve this question









$endgroup$




We have $$f(x,y,z)=xz+x^2z+sin(x+2y+z).$$ What is the the value of $df(1,-1,1)$.
I found the partial derivatives of f and than what? Is something like $$df(a,b,c,)=frac{df}{dx}(a,b,c)+frac{df}{dy}(a,b,c)+frac{df}{dz}(a,b,c)?$$
Thx guys.






partial-derivative




partial-derivative






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share|cite|improve this question










asked Dec 8 '18 at 11:35









NumbersNumbers

1426




1426












  • $begingroup$
    Not quite. Your left side, "$df(a,b,c)$" is a "differential" but the right side "$frac{partial f}{partial x}(a, b, c)+ frac{partial f}{partial y}(a, b, c)+ frac{partial f}{partial z}(a, b, c)$" is not. You want "$df(a,b,c)= frac{partial f}{partial x}(a, b, c)dx+ frac{partial f}{partial y}(a, b, c)dy+ frac{partial f}{partial z}(a, b, c)dz$".
    $endgroup$
    – user247327
    Dec 8 '18 at 11:47










  • $begingroup$
    Ok, I get it. Stil don t know the answer and how to find it.
    $endgroup$
    – Numbers
    Dec 8 '18 at 13:30


















  • $begingroup$
    Not quite. Your left side, "$df(a,b,c)$" is a "differential" but the right side "$frac{partial f}{partial x}(a, b, c)+ frac{partial f}{partial y}(a, b, c)+ frac{partial f}{partial z}(a, b, c)$" is not. You want "$df(a,b,c)= frac{partial f}{partial x}(a, b, c)dx+ frac{partial f}{partial y}(a, b, c)dy+ frac{partial f}{partial z}(a, b, c)dz$".
    $endgroup$
    – user247327
    Dec 8 '18 at 11:47










  • $begingroup$
    Ok, I get it. Stil don t know the answer and how to find it.
    $endgroup$
    – Numbers
    Dec 8 '18 at 13:30
















$begingroup$
Not quite. Your left side, "$df(a,b,c)$" is a "differential" but the right side "$frac{partial f}{partial x}(a, b, c)+ frac{partial f}{partial y}(a, b, c)+ frac{partial f}{partial z}(a, b, c)$" is not. You want "$df(a,b,c)= frac{partial f}{partial x}(a, b, c)dx+ frac{partial f}{partial y}(a, b, c)dy+ frac{partial f}{partial z}(a, b, c)dz$".
$endgroup$
– user247327
Dec 8 '18 at 11:47




$begingroup$
Not quite. Your left side, "$df(a,b,c)$" is a "differential" but the right side "$frac{partial f}{partial x}(a, b, c)+ frac{partial f}{partial y}(a, b, c)+ frac{partial f}{partial z}(a, b, c)$" is not. You want "$df(a,b,c)= frac{partial f}{partial x}(a, b, c)dx+ frac{partial f}{partial y}(a, b, c)dy+ frac{partial f}{partial z}(a, b, c)dz$".
$endgroup$
– user247327
Dec 8 '18 at 11:47












$begingroup$
Ok, I get it. Stil don t know the answer and how to find it.
$endgroup$
– Numbers
Dec 8 '18 at 13:30




$begingroup$
Ok, I get it. Stil don t know the answer and how to find it.
$endgroup$
– Numbers
Dec 8 '18 at 13:30










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You are given that $f(x,y,z)= xz+ x^2z+ sin(x+ 2y+ z)$



$frac{partial f}{partial x}= z+ 2xz+ cos(x+ 2y+ z)$



$frac{partial f}{partial x}(1,-1, 1)= 1+ 2+ cos(0)= 4$



$frac{partial f}{partial y}= 2cos(x+ 2y+ z)$



$frac{partial f}{partial y}(1,-1,1)= 2 cos(0)= 2$



$frac{partial f}{partial z}= x+ x^2+ cos(x+ 2y+ z)$



$frac{partial f}{partial z}(1,-1,1)= 1+ 1+ cos(0)= 3$



So $df= 4dx+ 2dy+ 3dz$.






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    $begingroup$

    You are given that $f(x,y,z)= xz+ x^2z+ sin(x+ 2y+ z)$



    $frac{partial f}{partial x}= z+ 2xz+ cos(x+ 2y+ z)$



    $frac{partial f}{partial x}(1,-1, 1)= 1+ 2+ cos(0)= 4$



    $frac{partial f}{partial y}= 2cos(x+ 2y+ z)$



    $frac{partial f}{partial y}(1,-1,1)= 2 cos(0)= 2$



    $frac{partial f}{partial z}= x+ x^2+ cos(x+ 2y+ z)$



    $frac{partial f}{partial z}(1,-1,1)= 1+ 1+ cos(0)= 3$



    So $df= 4dx+ 2dy+ 3dz$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You are given that $f(x,y,z)= xz+ x^2z+ sin(x+ 2y+ z)$



      $frac{partial f}{partial x}= z+ 2xz+ cos(x+ 2y+ z)$



      $frac{partial f}{partial x}(1,-1, 1)= 1+ 2+ cos(0)= 4$



      $frac{partial f}{partial y}= 2cos(x+ 2y+ z)$



      $frac{partial f}{partial y}(1,-1,1)= 2 cos(0)= 2$



      $frac{partial f}{partial z}= x+ x^2+ cos(x+ 2y+ z)$



      $frac{partial f}{partial z}(1,-1,1)= 1+ 1+ cos(0)= 3$



      So $df= 4dx+ 2dy+ 3dz$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You are given that $f(x,y,z)= xz+ x^2z+ sin(x+ 2y+ z)$



        $frac{partial f}{partial x}= z+ 2xz+ cos(x+ 2y+ z)$



        $frac{partial f}{partial x}(1,-1, 1)= 1+ 2+ cos(0)= 4$



        $frac{partial f}{partial y}= 2cos(x+ 2y+ z)$



        $frac{partial f}{partial y}(1,-1,1)= 2 cos(0)= 2$



        $frac{partial f}{partial z}= x+ x^2+ cos(x+ 2y+ z)$



        $frac{partial f}{partial z}(1,-1,1)= 1+ 1+ cos(0)= 3$



        So $df= 4dx+ 2dy+ 3dz$.






        share|cite|improve this answer









        $endgroup$



        You are given that $f(x,y,z)= xz+ x^2z+ sin(x+ 2y+ z)$



        $frac{partial f}{partial x}= z+ 2xz+ cos(x+ 2y+ z)$



        $frac{partial f}{partial x}(1,-1, 1)= 1+ 2+ cos(0)= 4$



        $frac{partial f}{partial y}= 2cos(x+ 2y+ z)$



        $frac{partial f}{partial y}(1,-1,1)= 2 cos(0)= 2$



        $frac{partial f}{partial z}= x+ x^2+ cos(x+ 2y+ z)$



        $frac{partial f}{partial z}(1,-1,1)= 1+ 1+ cos(0)= 3$



        So $df= 4dx+ 2dy+ 3dz$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 13:53









        user247327user247327

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