Value of multivariable function

Question

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.

Answer 1

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