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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.

Numbers
  • 539
  • 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 "$df(a,b,c)= \frac{\partial f}{\partial x}(a, b, c)dx+ \frac{\partial f}{\partial y}(a, b, c)dy+ \frac{\partial f}{\partial z}(a, b, c)dz$". – user247327 Dec 08 '18 at 11:47
  • Ok, I get it. Stil don t know the answer and how to find it. – Numbers Dec 08 '18 at 13:30

1 Answers1

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

user247327
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