Math Insight

Multivariable chain rule examples

The following are examples of using the multivariable chain rule. For examples involving the one-variable chain rule, see simple examples of using the chain rule or the chain rule from the Calculus Refresher.

Example 1

Let $\vc{g}: \R \rightarrow \R^2$ and $f: \R^2 \rightarrow \R$ (confused?) be defined by \begin{align*} \vc{g}(t) &= (t^3, t^4)\\ f(x,y) &= x^2y. \end{align*} (You can think of this as the mountain climbing example where $f(x,y)$ is height of mountain at point $(x,y)$ and the path $\vc{g}(t)$ gives your position at time $t$.) Let $h(t)$ be the composition of $f$ with $\vc{g}$ (which would give your height at time $t$): \begin{align*} h(t) = (f \circ \vc{g}) (t) = f(\vc{g}(t)). \end{align*} Calculate the derivative $\displaystyle h'(t) = \diff{h}{t}(t)$ (i.e., the change in height) via the chain rule.

Solution A: We'll use the formula using matrices of partial derivatives: \begin{align*} D{h}(t) = Df(\vc{g}(t)) D{\vc{g}}(t). \end{align*}

We calculate the matrices of partial derivatives of $f$ and $\vc{g}$. \begin{align*} Df(x,y) &= \left[ \pdiff{f}{x}(x,y) \quad \pdiff{f}{y}(x,y) \right]\\ &=\left[ \begin{array}{cc} 2xy & x^2 \end{array} \right]\\ D\vc{g}(t) &= \left[ \begin{array}{c} g_1'(t) \\g_2'(t) \end{array} \right] = \left[ \begin{array}{c} 3t^2\\4t^3 \end{array} \right] \end{align*} We need to evaluate $Df$ at the point $\vc{g}(t)$: \begin{align*} Df(\vc{g}(t)) =Df(t^3, t^4)= \left[ \begin{array}{cc} 2(t^3)(t^4) & (t^3)^2 \end{array} \right] = \left[ \begin{array}{cc} 2t^7 & t^6 \end{array} \right] \end{align*} The derivative of $h$ is \begin{align*} h'(t)=\diff{h}{t}(t) = Dh(t) &= Df(\vc{g}(t)) D\vc{g}(t)\\ &= \left[ \begin{array}{cc} 2t^7 & t^6 \end{array} \right] \left[ \begin{array}{c} 3t^2\\4t^3 \end{array} \right]\\ &= (2t^7)(3t^2) + (t^6)(4t^3)= 6 t^9 + 4 t^9\\ &= 10 t^9 \end{align*}

Solution B: We'll start immediately with the formula in component form: \begin{align*} \diff{h}{t}(t) = \pdiff{f}{x}(\vc{g}(t)) \diff{g_1}{t} (t) + \pdiff{f}{y}(\vc{g}(t))\diff{g_2}{t} (t). \end{align*} We calculate \begin{align*} \pdiff{f}{x}(x,y) &= 2xy\\ \pdiff{f}{y}(x,y) &= x^2\\ \pdiff{f}{x}(\vc{g}(t)) &= \pdiff{f}{x}(t^3,t^4) = 2(t^3)(t^4) = 2t^7\\ \pdiff{f}{y}(\vc{g}(t)) &= \pdiff{f}{y}(t^3,t^4) = (t^3)^2 = t^6\\ \diff{g_1}{t}(t) &= 3t^2\\ \diff{g_2}{t}(t) &= 4t^3. \end{align*} Therefore, \begin{align*} \diff{h}{t}(t) = (2t^7)(3t^2) + (t^6)(4t^3) =6 t^9 + 4 t^9 = 10 t^9. \end{align*}

Example 1'

Verify the chain rule for example 1 by calculating an expression for $h(t)$ and then differentiating it to obtain $\displaystyle \diff{h}{t}(t)$.

Solution: $h(t) = f(\vc{g}(t)) = f(t^3,t^4) = (t^3)^2(t^4) = t^{10}$. \begin{align*} h'(t) = \diff{h}{t}(t) = 10t^9, \end{align*} which matches the solution to Example 1, verifying that the chain rule got the correct answer.

For this simple example, doing it without the chain rule was a lot easier. However, that is not always the case. And, in the next example, the only way to obtain the answer is to use the chain rule.

Example 2

We continue the mountain climbing example of Example 1. But now, let's say we don't know the terrain ahead of time. This means we do not yet know the height $f(x,y)$ at the position $(x,y)$. We do, however, know our path through mountain; as before, it is given by $\vc{g}(t) = (t^3, t^4).$

Calculate the change in height that you'll experience along the path, i.e., calculate the derivative of $h(t) = f(\vc{g}(t))$. In this case, since we don't know $f$, the answer will be given in terms of the function $f(x,y)$.

Solution: We'll just copy solution A, above. This time, though, we must leave the matrix of partial derivatives of $f$ as \begin{align*} Df(x,y)= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(x,y) & \displaystyle \pdiff{f}{y}(x,y) \end{array} \right] \end{align*} since we don't know what $f(x,y)$ is. We can substitute in the values along the path $\vc{g}(t)$: \begin{align*} Df(\vc{g}(t)) = Df(t^3,t^4)= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(t^3,t^4) & \displaystyle \pdiff{f}{y}(t^3,t^4) \end{array} \right]. \end{align*} Since $D\vc{g}(t)$ is the same as in solution A, above, we calculate the derivative of $h$ as \begin{align*} h'(t) = Dh(t) &= Df(\vc{g}(t)) D\vc{g}(t)\\ &= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(t^3,t^4) & \displaystyle \pdiff{f}{y}(t^3,t^4) \end{array} \right] \left[ \begin{array}{c} 3t^2\\4t^3 \end{array} \right]\\ &=3t^2\pdiff{f}{x}(t^3,t^4) +4t^3\pdiff{f}{y}(t^3,t^4). \end{align*} We leave the answer in this form. Of course, as soon as we know what $f(x,y)$ is, we can simply compute its partial derivatives and plug the result into this formula.

Example 3

We continue using the same function $f(x,y) = x^2y$ to describe the height of the mountain at position $(x,y)$. We embellish the above examples by letting $g: \R^2 \to \R^2$ be defined by $\vc{g}(s,t) = (t-s^2,ts^2)$. (We could think of having many paths through the mountain that depend on a skill level $s$. Then, $(x,y)=\vc{g}(s,t)$ could be the position of a person at time $t$ with skill level $s$.)

Compute $\pdiff{}{s} (f \circ \vc{g})(s,t)$ and $\pdiff{}{t} (f \circ \vc{g})(s,t)$, i.e., the partial derivatives with respect to $s$ and $t$ of the height of a person in the mountains whose position is given by $\vc{g}(s,t)$.

Solution: Let $h(s,t) = (f \circ \vc{g})(s,t) = f(\vc{g}(s,t))$. We need to calculate $\displaystyle \pdiff{h}{s}(s,t)$ and $\displaystyle \pdiff{h}{t}(s,t)$. The chain rule says that \begin{align*} Dh(s,t) = D(f\circ \vc{g})(s,t) = Df(\vc{g}(s,t)) D\vc{g}(s,t). \end{align*} Since \begin{align*} Dh(s,t) = \left[ \pdiff{h}{s}(s,t) \quad \pdiff{h}{t}(s,t)\right], \end{align*} the answers we want are just the two components of $Dh(s,t)$. We just need to calculate the matrices $Df(\vc{g}(s,t))$ and $D\vc{g}(s,t)$, then multiply them together.

To make it easier in case you have to do such a problem again, we'll perform the matrix multiplication before writing in the specific values for $f(x,y)$ and $\vc{g}(s,t)$. Then, we'll end up with the chain rule written in component form, which may be easier to use.

The function $f(x,y)$ hasn't changed, so its matrix of partial derivatives is \begin{align*} Df(x,y)= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(x,y) & \displaystyle \pdiff{f}{y}(x,y) \end{array} \right]. \end{align*} For the chain rule, we need this evaluated at $(x,y)=\vc{g}(s,t)$ \begin{align*} Df(\vc{g}(s,t))= \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}(\vc{g}(s,t)) & \displaystyle \pdiff{f}{y}(\vc{g}(s,t)) \end{array} \right]. \end{align*} Since $\vc{g}: \R^2 \to \R^2$, its matrix of partial derivatives is a $2 \times 2$ matrix. If we denote its components as $\vc{g}(s,t) = (g_1(s,t), g_2(s,t))$, its matrix of partial derivatives is \begin{align*} D\vc{g}(s,t)= \left[ \begin{array}{cc} \displaystyle\pdiff{g_1}{s}(s,t)& \displaystyle\pdiff{g_1}{t}(s,t)\\ \displaystyle\pdiff{g_2}{s}(s,t)& \displaystyle\pdiff{g_2}{t}(s,t) \end{array} \right]. \end{align*} The chain rule $Dh(s,t) = Df(\vc{g}(s,t)) D\vc{g}(s,t)$ becomes \begin{align*} \left[ \begin{array}{cc} \displaystyle \pdiff{h}{s}(s,t) & \displaystyle \pdiff{h}{t} (s,t) \end{array} \right] = \left[ \begin{array}{cc} \displaystyle \pdiff{f}{x}\bigl(\vc{g}(s,t)\bigr) & \displaystyle \pdiff{f}{y}\bigl(\vc{g}(s,t)\bigr) \end{array} \right] \left[ \begin{array}{cc} \displaystyle \pdiff{g_1}{s}(s,t) & \displaystyle \pdiff{g_1}{t}(s,t) \\ \displaystyle \pdiff{g_2}{s}(s,t) & \displaystyle \pdiff{g_2}{t}(s,t) \end{array} \right] \end{align*} We can compute the matrix product on the right-hand side; the result is a $1 \times 2$ matrix (i.e., the same size of $Dh(s,t)$). We obtain one equation by matching the first component of $Dh(s,t)$ with the first component of this multiplied-out matrix. We obtain a second equation by matching the second component of $Dh(s,t)$ with the second component of this multiplied-out matrix. The resulting two equations are \begin{align*} \pdiff{h}{s}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{s}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{s}(s,t)\\ \pdiff{h}{t}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{t}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{t}(s,t). \end{align*} This is the chain rule written out in component form for $h : \R^2 \to \R$, $f : \R^2 \to \R$, and $\vc{g} : \R^2 \to \R^2$. It is equation () from the special case page.

Now, we compute the answer to our specific problem by substituting in for $f(x,y) = x^2y$ and $\vc{g}(s,t) = (t-s^2, ts^2)$. \begin{align*} \pdiff{f}{x}(x,y) &= 2xy &\pdiff{f}{x}(\vc{g}(s,t)) &= 2ts^2(t-s^2)\\ \pdiff{f}{y}(x,y) &= x^2 & \pdiff{f}{y}(\vc{g}(s,t)) &= (t-s^2)^2\\ \pdiff{g_1}{s}(s,t) &= -2s &\pdiff{g_2}{s}(s,t) &= 2st\\ \pdiff{g_1}{t}(s,t) &= 1 &\pdiff{g_2}{t}(s,t) &= s^2 \end{align*} Finally, we get our answers. \begin{align*} \pdiff{}{s} (f \circ \vc{g})(s,t) = \pdiff{h}{s}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{s}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{s}(s,t)\\ &= 2ts^2(t-s^2)(-2s) + (t-s^2)^2(2st)\\ &= -4ts^3(t-s^2) + 2st(t^2-2ts^2+s^4)\\ &= -4s^3t^2 + 4s^5t +2st^3 -4s^3t^2 + 2s^5t\\ &= -8s^3t^2+ 6s^5t+2st^3 \end{align*} \begin{align*} \pdiff{}{t} (f \circ \vc{g})(s,t) = \pdiff{h}{t}(s,t) &= \pdiff{f}{x}(\vc{g}(s,t))\pdiff{g_1}{t}(s,t) + \pdiff{f}{y}(\vc{g}(s,t))\pdiff{g_2}{t}(s,t)\\ &= 2ts^2(t-s^2)(1) + (t-s^2)^2(s^2)\\ &= 2s^2t^2 - 2s^4t + s^2t^2 -2s^4t + s^6\\ &= 3s^2t^2- 4s^4t+ s^6 \end{align*}