# Chain rule with multivariable functions

Let $u(x,y)=x+y^2, v(x,y)=y^2 x$ and $h(x,y)=f(g(u(x,y),v(x,y)))$, where $f,g$ are differentiables.

Further $g(0,-1)=4,\frac{df(4)}{dx}=2,\frac{\partial h(-1,1)}{\partial y}=\frac{\partial h(-1,1)}{\partial x}=-8$.

Find the equation of the plane tangent to the graph of $g$ at the point $(0,-1,4)$.

I'm confused about how to apply the chain rule in the $h(x,y)$ function.

I'm also a student studying this but I'll give it my best shot. The function $h$ might be visually explained by the following:

   f
|
g
/ \
u v
/| |\
x y x y


When differentiating a bunch of nested functions as above, you use the single-variable chain rule for each function in the diagram with one "child" and the multivariable chain rule for each function with multiple children. We basically recursively go down this diagram, applying the correct chain rule to $f$, then $g$, and so forth, if there were more nested functions.

Since we are finding the tangent plane, we'll need two vectors. If you can get the $x$- and $y$-partial derivatives of $g$, then you'll pretty much be done — the vectors spanning the plane will be $\langle 1, 0, g_x(0, -1) \rangle$ and $\langle 0, 1, g_y(0, -1) \rangle$.

Applying the single-variable chain rule to $f(x)$, we get

$$h_x(x, y) = f'(g(u(x,y),v(x,y))){\partial \over \partial x}\left[g(u(x,y),v(x,y))\right]$$

which, after applying the multivariable chain rule to $g$, becomes

$$f'(g(u(x,y),v(x,y)))\left[g_x(u(x,y),v(x,y))u_x(x,y) + g_y(u(x,y),v(x,y))v_x(x,y)\right].$$

Looks hefty, but we can simplify. Notice they gave you $g$ evaluated at $(0, -1)$, so we want to pick $x$ and $y$ that will give $u(x, y) = 0$ and $v(x, y) = -1$. $(-1, 1)$ will do the trick, as indicated by the input they gave you for $\partial h \over \partial x$.

$$u_x(x, y) = 1; v_x(x, y) = y^2 = 1$$ $$u(-1, 1) = 0; v(-1, 1) = -1$$ $$h_x(-1, 1) = f'(g(0,-1))\left[g_x(0,-1)u_x(-1,1) + g_y(0,-1))v_x(-1,1)\right]$$ $$-8 = f'(4)\left[g_x(0,-1) + g_y(0,-1)\right]$$ $$-4 = g_x(0,-1) + g_y(0,-1)$$

Now do the same by partially deriving with respect to $y$. You'll have a linear system of two equations with $g_x$ and $g_y$ unknown, which can be solved by eliminating one of the variables and plugging back in.

For the final answer, the vectors spanning the plane will then be $\langle 1, 0, g_x(0, -1) \rangle$ and $\langle 0, 1, g_y(0, -1) \rangle$. So the equation of the plane will be of the form

$$p(s, t) = \langle 0, -1, 4 \rangle + s\langle 1, 0, g_x(0, -1) \rangle + t\langle 0, 1, g_y(0, -1) \rangle$$

• Nicely done. I would’ve derived an implicit Cartesian equation for the plane using the normal $(g_x,g_y,-1)$ instead (or in addition). Note that you’ve also been given the value of $f'(4)$ in the problem.
– amd
Jul 18, 2018 at 1:09
• Thanks for pointing that out! I've updated my answer to reflect that $f'(4)$ is known. Jul 18, 2018 at 23:35

I hate this notation for partial derivatives. Here’s how I would go about solving this.

First, introduce an auxiliary function $\phi:\mathbb R^2\to\mathbb R^2$ with $$\phi: \begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}u(x,y)\\v(x,y)\end{bmatrix} = \begin{bmatrix} x+y^2 \\ y^2x \end{bmatrix}.$$ We then have $h = f\circ g\circ \phi$ and applying the chain rule twice gives $$\mathrm dh_{(x,y)} = \mathrm df_{g(\phi(x,y))}\circ\mathrm dg_{\phi(x,y)}\circ\mathrm d\phi_{(x,y)}.$$ Expanding by coordinates, this becomes the matrix product \begin{align} \begin{bmatrix}\partial_1 h(x,y)&\partial_2 h(x,y)\end{bmatrix} &= f'(g(\phi(x,y))) \begin{bmatrix}\partial_1g(\phi(x,y)) & \partial_2g(\phi(x,y))\end{bmatrix} \begin{bmatrix} \partial_1u(x,y) & \partial_2u(x,y) \\ \partial_1v(x,y) & \partial_2v(x,y)\end{bmatrix} \\ &= f'(g(\phi(x,y))) \begin{bmatrix}\partial_1g(\phi(x,y)) & \partial_2g(\phi(x,y))\end{bmatrix} \begin{bmatrix} 1 & 2y \\ y^2 & 2xy\end{bmatrix}. \end{align} We’re given that $g(0,-1)=4$, $f'(4)=2$, and $\partial_1h(-1,1)=\partial_2h(-1,1)=-8.$ Since $g$ is supposedly differentiable, we’ll hand-wave away the fact that the system of equations $u(x,y)=0$, $v(x,y)=-1$ actually has two solutions that lead to different values for $\mathrm dg(0,-1)$ and assume that the domains of $u$ and $v$ are suitably resticted so that we can take $x=-1$ and $y=1$, as appears to have been intended. Plugging these known values into the above equation produces, with a slight abuse of notation, $$\begin{bmatrix}-8&-8\end{bmatrix} = 2 \begin{bmatrix}\partial_1g(0,-1) & \partial_2g(0,-1)\end{bmatrix} \begin{bmatrix} 1 & 2 \\ 1 & -2\end{bmatrix}$$ or \begin{align}\partial_1g(0,-1)+\partial_2g(0,-1) &= -4 \\ 2\partial_1g(0,-1)-2\partial_2g(0,-1) &= -4.\end{align} Solving this simple system of linear equations gives you the values of the partial derivatives of $g$ at $(0,-1)$, from which I assume you can construct the tangent plane.