## Thomas' Calculus 13th Edition

$\nabla f =-i+j$ starting at initial point (2,1) on the level curve $-1 =\space y-x$
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant to find the x-ordinate of the gradient vector, and vice versa: $f_x(2,1)=\dfrac{\partial }{\partial x}(y-x)=-x^{0}=-1|_{(2,1)}$ and $f_y(2,1)=\dfrac{\partial }{\partial x}(y-x)=y^{0}=1|_{(2,1)}$ Write the gradient vector that extends 1 unit to the left and 1 unit up. $\nabla f = \lt -1,1 \gt =-i+j$ and $f(2,1)=1-2=-1$ So, the equation of the level curve is: $-1 =\space y-x$ Thus, we have: $\nabla f =-i+j$ starting at initial point (2,1) on the level curve $-1 =\space y-x$