Multivariable calculus and vector analysis

Chain rule examples

Example 1

Let $\mathbf{g}:\mathbf{R}\to {\mathbf{R}}^2$ and $f:{\mathbf{R}}^2\to \mathbf{R}$ be defined by

$$\begin{array}{lll}\hfill \mathbf{g}\left(t\right)=\left(t^3,t^4\right)& & \hfill \\ \hfill f\left(x,y\right)=x^{2}y.& & \hfill \end{array}$$

(You can think of this as the mountain climbing example where $f\left(x,y\right)$ is height of mountain at point $\left(x,y\right)$ and the path $\mathbf{g}\left(t\right)$ gives your position at time $t$.) Let $h\left(t\right)$ be the composition of $f$ with $\mathbf{g}$ (which would give your height at time $t$):

$$\begin{array}{lll}\hfill h\left(t\right)=\left(f\circ \mathbf{g}\right)\left(t\right)=f\left(\mathbf{g}\left(t\right)\right)& & \hfill \end{array}$$

Calculate the derivative $h^{\text{'}}\left(t\right)=\frac{\mathrm{d}h}{\mathrm{d}t}\left(t\right)$ (i.e., the change in height) via the chain rule.

Solution A: We’ll use the formula using matrices of partial derivatives:

$$\begin{array}{lll}\hfill Dh\left(t\right)=Df\left(\mathbf{g}\left(t\right)\right)D\mathbf{g}\left(t\right).& & \hfill \end{array}$$

We calculate the matrices of partial derivatives of $f$ and $\mathbf{g}$.

$$\begin{array}{lll}\hfill Df\left(x,y\right)=\left[\frac{\partial f}{\partial x}\left(x,y\right)\frac{\partial f}{\partial y}\left(x,y\right)\right]& & \hfill \\ \hfill =\left[\begin{array}{cc}\hfill 2xy\hfill & \hfill x^2\hfill \end{array}\right]& & \hfill \\ \hfill D\mathbf{g}\left(t\right)=\left[\begin{array}{c}\hfill g_1^{\text{'}}\left(t\right)\hfill \\ \hfill g_2^{\text{'}}\left(t\right)\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 3t^2\hfill \\ \hfill 4t^3\hfill \end{array}\right]& & \hfill \end{array}$$

We need to evaluate $Df$ at the point $\mathbf{g}\left(t\right)$:

$$\begin{array}{lll}\hfill Df\left(\mathbf{g}\left(t\right)\right)=Df\left(t^3,t^4\right)=\left[\begin{array}{cc}\hfill 2\left(t^3\right)\left(t^4\right)\hfill & \hfill {\left(t^3\right)}^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2t^7\hfill & \hfill t^6\hfill \end{array}\right]& & \hfill \end{array}$$

The derivative of $h$ is

$$\begin{array}{llll}\hfill h^{\text{'}}\left(t\right)=\frac{\mathrm{d}h}{\mathrm{d}t}\left(t\right)=Dh\left(t\right)& =Df\left(\mathbf{g}\left(t\right)\right)D\mathbf{g}\left(t\right)& \hfill & \\ \hfill & =\left[\begin{array}{cc}\hfill 2t^7\hfill & \hfill t^6\hfill \end{array}\right]\left[\begin{array}{c}\hfill 3t^2\hfill \\ \hfill 4t^3\hfill \end{array}\right]& \hfill & \\ \hfill & =\left(2t^7\right)\left(3t^2\right)+\left(t^6\right)\left(4t^3\right)=6t^9+4t^9& \hfill & \\ \hfill & =10t^9& \hfill & \end{array}$$

Solution B: We’ll start immediately with the formula in component form:

$$\begin{array}{lll}\hfill \frac{\mathrm{d}h}{\mathrm{d}t}\left(t\right)=\frac{\partial f}{\partial x}\left(\mathbf{g}\left(t\right)\right)\frac{\mathrm{d}{g}_1}{\mathrm{d}t}\left(t\right)+\frac{\partial f}{\partial y}\left(\mathbf{g}\left(t\right)\right)\frac{\mathrm{d}{g}_2}{\mathrm{d}t}\left(t\right).& & \hfill \end{array}$$

We calculate

$$\begin{array}{llll}\hfill \frac{\partial f}{\partial x}\left(x,y\right)& =2xy& \hfill & \\ \hfill \frac{\partial f}{\partial y}\left(x,y\right)& =x^2& \hfill & \\ \hfill \frac{\partial f}{\partial x}\left(\mathbf{g}\left(t\right)\right)& =\frac{\partial f}{\partial x}\left(t^3,t^4\right)=2\left(t^3\right)\left(t^4\right)=2t^7& \hfill & \\ \hfill \frac{\partial f}{\partial y}\left(\mathbf{g}\left(t\right)\right)& =\frac{\partial f}{\partial y}\left(t^3,t^4\right)={\left(t^3\right)}^2=t^6& \hfill & \\ \hfill \frac{\mathrm{d}{g}_1}{\mathrm{d}t}\left(t\right)& =3t^2& \hfill & \\ \hfill \frac{\mathrm{d}{g}_2}{\mathrm{d}t}\left(t\right)& =4t^3.& \hfill & \end{array}$$

Therefore,

$$\begin{array}{lll}\hfill \frac{\mathrm{d}h}{\mathrm{d}t}\left(t\right)=\left(2t^7\right)\left(3t^2\right)+\left(t^6\right)\left(4t^3\right)=6t^9+4t^9=10t^9.& & \hfill \end{array}$$

Example 1’

Verify the chain rule for example 1 by calculating an expression for $h\left(t\right)$ and then differentiating it to obtain $\frac{\mathrm{d}h}{\mathrm{d}t}\left(t\right)$.

Solution: $h\left(t\right)=f\left(\mathbf{g}\left(t\right)\right)=f\left(t^3,t^4\right)={\left(t^3\right)}^2\left(t^4\right)=t^{10}$.

$$\begin{array}{lll}\hfill h^{\text{'}}\left(t\right)=\frac{\mathrm{d}h}{\mathrm{d}t}\left(t\right)=10t^9,& & \hfill \end{array}$$

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 $f\left(x,y\right)$ at the position $\left(x,y\right)$. We do, however, know our path through mountain; as before, it is given by $\mathbf{g}\left(t\right)=\left(t^3,t^4\right).$

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

Solution: We’ll just copy solution A, above. This time, though, we must leave the matrix of partial derivatives of $f$ as

$$\begin{array}{lll}\hfill Df\left(x,y\right)=\left[\begin{array}{cc}\hfill \frac{\partial f}{\partial x}\left(x,y\right)\hfill & \hfill \frac{\partial f}{\partial y}\left(x,y\right)\hfill \end{array}\right]& & \hfill \end{array}$$

since we don’t know what $f\left(x,y\right)$ is. We can substitute in the values along the path $\mathbf{g}\left(t\right)$:

$$\begin{array}{lll}\hfill Df\left(\mathbf{g}\left(t\right)\right)=Df\left(t^3,t^4\right)=\left[\begin{array}{cc}\hfill \frac{\partial f}{\partial x}\left(t^3,t^4\right)\hfill & \hfill \frac{\partial f}{\partial y}\left(t^3,t^4\right)\hfill \end{array}\right].& & \hfill \end{array}$$

Since $D\mathbf{g}\left(t\right)$ is the same as in solution A, above, we calculate the derivative of $h$ as

$$\begin{array}{llll}\hfill h^{\text{'}}\left(t\right)=Dh\left(t\right)& =Df\left(\mathbf{g}\left(t\right)\right)D\mathbf{g}\left(t\right)& \hfill & \\ \hfill & =\left[\begin{array}{cc}\hfill \frac{\partial f}{\partial x}\left(t^3,t^4\right)\hfill & \hfill \frac{\partial f}{\partial y}\left(t^3,t^4\right)\hfill \end{array}\right]\left[\begin{array}{c}\hfill 3t^2\hfill \\ \hfill 4t^3\hfill \end{array}\right]& \hfill & \\ \hfill & =3t^2\frac{\partial f}{\partial x}\left(t^3,t^4\right)+4t^3\frac{\partial f}{\partial y}\left(t^3,t^4\right).& \hfill & \end{array}$$

We leave the answer in this form. Of course, as soon as we know what $f\left(x,y\right)$ is, we can simply compute its partial derivatives and plug the result into this formula.

Example 3

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

Compute $\frac{\partial }{\partial s}\left(f\circ \mathbf{g}\right)\left(s,t\right)$ and $\frac{\partial }{\partial t}\left(f\circ \mathbf{g}\right)\left(s,t\right)$, 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 $\mathbf{g}\left(s,t\right)$.

Solution: Let $h\left(s,t\right)=\left(f\circ \mathbf{g}\right)\left(s,t\right)=f\left(\mathbf{g}\left(s,t\right)\right)$. We need to calculate $\frac{\partial h}{\partial s}\left(s,t\right)$ and $\frac{\partial h}{\partial t}\left(s,t\right)$. The chain rule says that

$$\begin{array}{lll}\hfill Dh\left(s,t\right)=D\left(f\circ \mathbf{g}\right)\left(s,t\right)=Df\left(\mathbf{g}\left(s,t\right)\right)D\mathbf{g}\left(s,t\right).& & \hfill \end{array}$$

Since

$$\begin{array}{lll}\hfill Dh\left(s,t\right)=\left[\frac{\partial h}{\partial s}\left(s,t\right)\frac{\partial h}{\partial t}\left(s,t\right)\right],& & \hfill \end{array}$$

the answers we want are just the two components of $Dh\left(s,t\right)$. We just need to calculate the matrices $Df\left(\mathbf{g}\left(s,t\right)\right)$ and $D\mathbf{g}\left(s,t\right)$, 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 $f\left(x,y\right)$ and $\mathbf{g}\left(s,t\right)$. Then, we’ll end up with the chain rule written in component form, which may be easier to use.

The function $f\left(x,y\right)$ hasn’t changed, so its matrix of partial derivatives is

$$\begin{array}{lll}\hfill Df\left(x,y\right)=\left[\begin{array}{cc}\hfill \frac{\partial f}{\partial x}\left(x,y\right)\hfill & \hfill \frac{\partial f}{\partial y}\left(x,y\right)\hfill \end{array}\right].& & \hfill \end{array}$$

For the chain rule, we need this evaluated at $\left(x,y\right)=\mathbf{g}\left(s,t\right)$

$$\begin{array}{lll}\hfill Df\left(\mathbf{g}\left(s,t\right)\right)=\left[\begin{array}{cc}\hfill \frac{\partial f}{\partial x}\left(\mathbf{g}\left(s,t\right)\right)\hfill & \hfill \frac{\partial f}{\partial y}\left(\mathbf{g}\left(s,t\right)\right)\hfill \end{array}\right].& & \hfill \end{array}$$

Since $\mathbf{g}:{\mathbf{R}}^2\to {\mathbf{R}}^2$, its matrix of partial derivatives is a $2\times 2$ matrix. If we denote its components as $\mathbf{g}\left(s,t\right)=\left(g_1\left(s,t\right),g_2\left(s,t\right)\right)$, its matrix of partial derivatives is

$$\begin{array}{lll}\hfill D\mathbf{g}\left(s,t\right)=\left[\begin{array}{cc}\hfill \frac{\partial g_1}{\partial s}\left(s,t\right)\hfill & \hfill \frac{\partial g_1}{\partial t}\left(s,t\right)\hfill \\ \hfill \frac{\partial g_2}{\partial s}\left(s,t\right)\hfill & \hfill \frac{\partial g_2}{\partial t}\left(s,t\right)\hfill \end{array}\right].& & \hfill \end{array}$$

The chain rule $Dh\left(s,t\right)=Df\left(\mathbf{g}\left(s,t\right)\right)D\mathbf{g}\left(s,t\right)$ becomes

$$\begin{array}{lll}\hfill \left[\begin{array}{cc}\hfill \frac{\partial h}{\partial s}\left(s,t\right)\hfill & \hfill \frac{\partial h}{\partial t}\left(s,t\right)\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \frac{\partial f}{\partial x}\left(\mathbf{g}\left(s,t\right)\right)\hfill & \hfill \frac{\partial f}{\partial y}\left(\mathbf{g}\left(s,t\right)\right)\hfill \end{array}\right]\left[\begin{array}{cc}\hfill \frac{\partial g_1}{\partial s}\left(s,t\right)\hfill & \hfill \frac{\partial g_1}{\partial t}\left(s,t\right)\hfill \\ \hfill \frac{\partial g_2}{\partial s}\left(s,t\right)\hfill & \hfill \frac{\partial g_2}{\partial t}\left(s,t\right)\hfill \end{array}\right]& & \hfill \end{array}$$

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 $Dh\left(s,t\right)$). We obtain one equation by matching the first component of $Dh\left(s,t\right)$ with the first component of this multiplied-out matrix. We obtain a second equation by matching the second component of $Dh\left(s,t\right)$ with the second component of this multiplied-out matrix. The resulting two equations are

$$\begin{array}{llll}\hfill \frac{\partial h}{\partial s}\left(s,t\right)& =\frac{\partial f}{\partial x}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_1}{\partial s}\left(s,t\right)+\frac{\partial f}{\partial y}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_2}{\partial s}\left(s,t\right)& \hfill & \\ \hfill \frac{\partial h}{\partial t}\left(s,t\right)& =\frac{\partial f}{\partial x}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_1}{\partial t}\left(s,t\right)+\frac{\partial f}{\partial y}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_2}{\partial t}\left(s,t\right).& \hfill & \end{array}$$

This is the chain rule written out in component form for $h:{\mathbf{R}}^2\to \mathbf{R}$, $f:{\mathbf{R}}^2\to \mathbf{R}$, and $\mathbf{g}:{\mathbf{R}}^2\to {\mathbf{R}}^2$.

Now, we compute the answer to our specific problem by substituting in for $f\left(x,y\right)=x^{2}y$ and $\mathbf{g}\left(s,t\right)=\left(t-s^2,t{s}^2\right)$.

$$\begin{array}{llllllll}\hfill \frac{\partial f}{\partial x}\left(x,y\right)& =2xy& \hfill \frac{\partial f}{\partial x}\left(\mathbf{g}\left(s,t\right)\right)& =2t{s}^2\left(t-s^2\right)& \hfill & & \hfill & \\ \hfill \frac{\partial f}{\partial y}\left(x,y\right)& =x^2& \hfill \frac{\partial f}{\partial y}\left(\mathbf{g}\left(s,t\right)\right)& ={\left(t-s^2\right)}^2& \hfill & & \hfill & \\ \hfill \frac{\partial g_1}{\partial s}\left(s,t\right)& =-2s& \hfill \frac{\partial g_2}{\partial s}\left(s,t\right)& =2st& \hfill & & \hfill & \\ \hfill \frac{\partial g_1}{\partial t}\left(s,t\right)& =1& \hfill \frac{\partial g_2}{\partial t}\left(s,t\right)& =s^2& \hfill & & \hfill & \end{array}$$

Finally, we get our answers.

$$\begin{array}{llll}\hfill \frac{\partial }{\partial s}\left(f\circ \mathbf{g}\right)\left(s,t\right)=\frac{\partial h}{\partial s}\left(s,t\right)& =\frac{\partial f}{\partial x}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_1}{\partial s}\left(s,t\right)+\frac{\partial f}{\partial y}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_2}{\partial s}\left(s,t\right)& \hfill & \\ \hfill & =2t{s}^2\left(t-s^2\right)\left(-2s\right)+{\left(t-s^2\right)}^2\left(2st\right)& \hfill & \\ \hfill & =-4t{s}^3\left(t-s^2\right)+2st\left(t-2t{s}^2+s^4\right)& \hfill & \\ \hfill & =-4s^3{t}^2+4s^{5}t+2s{t}^2-4s^3{t}^2+2s^{5}t& \hfill & \\ \hfill & =-8s^3{t}^2+6s^{5}t+2s{t}^2& \hfill & \end{array}$$ $$\begin{array}{llll}\hfill \frac{\partial }{\partial t}\left(f\circ \mathbf{g}\right)\left(s,t\right)=\frac{\partial h}{\partial t}\left(s,t\right)& =\frac{\partial f}{\partial x}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_1}{\partial t}\left(s,t\right)+\frac{\partial f}{\partial y}\left(\mathbf{g}\left(s,t\right)\right)\frac{\partial g_2}{\partial t}\left(s,t\right)& \hfill & \\ \hfill & =2t{s}^2\left(t-s^2\right)\left(1\right)+{\left(t-s^2\right)}^2\left(s^2\right)& \hfill & \\ \hfill & =2s^2{t}^2-2s^{4}t+s^2{t}^2-2s^{4}t+s^6& \hfill & \\ \hfill & =3s^2{t}^2-4s^{4}t+s^6& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

2a. The partial derivative

2b. The derivative and differentiability

2c. The chain rule, directional derivative and gradient

• The chain rule
• Chain rule examples
• The directional derivative and the gradient
• Directional derivative and gradient examples

2d. Applications of differential calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material