Answer
$ \lt 3,2,-4 \gt $
Work Step by Step
Our aim is to take the first partial derivative of the given function $f(x,y,z)$ with respect to $x$, by treating $y$ and $z$ as a constant, and vice versa:
$f_x=2x+\dfrac{z}{x}$
$f_y=2y \implies f_y(2,1)=\dfrac{\partial }{\partial x}(y-x)=y^{0}=1$
$f_z= -4z+\ln x$
The gradient vector equation is: $\nabla f = \lt f_x,f_y,f_z \gt $
$\nabla f (1,1,1) = \lt 2x+\dfrac{z}{x},2y , -4z+\ln x\gt $
or, $ \lt 2(1)+\dfrac{1}{1},2(1) , -4(1)+\ln (1) \gt = \lt 3,2,-4 \gt $
