# 14.2 | Computing Partial Derivatives Algebraically

Consider the surface defined as the level set $$x^3+y^3+z^3+4xyz = 0$$, and $$P(1,-1,2)$$ a point on this surface. Assume that close to $$P$$ the surface determines an implicit function $$z = h(x,y)$$.

1. Determine the gradient of $$h$$ at $$P$$.
2. Determine a direction $$P$$ (a direction with respect to $$x$$ and $$y$$) such that the rate of change of $$h$$ is zero.

1. \begin{aligned} \frac{d}{dx}[x^3 +y^3 + z^3 + 4xyz] & = \frac{d}{dx}[0] \\ \frac{d}{dx}[x^3] +\frac{d}{dx}[y^3] + \frac{d}{dx}[z^3] + \frac{d}{dx}[4xyz] & = 0 \\ 3x^2 + 3z^2\frac{dz}{dx}+4xy\frac{dz}{dx} & = 0 \\ \frac{dz}{dx}(3z^2+4xy) & = -3x^2-4yz \\ \frac{dz}{dx} & = \frac{-3x^2-4yz}{3z^2+4xy} \end{aligned}
or
\begin{aligned} \frac{dz}{dx} & = -\frac{\frac{dF}{dx}}{\frac{dF}{dz}} \\ & = -\frac{3x^2+4yz}{3z^2+4xy} \end{aligned}
where $$F(x,y,z) = x^3 +y^3 + z^3 + 4xyz$$. So,
$$h_x = -\frac{3x^2+4yz}{3z^2+4xy}.$$
Similarly,
$$h_y = -\frac{3y^2 + 4xz}{3z^2+4xy}.$$
So, $$\nabla h = \frac{1}{3z^2+4xy} \langle 3x^2+4yz, 3y^2+4xz \rangle$$ and $$\nabla h(1,-1,2) = \langle 5/8,-11/5 \rangle$$.