3.6 Chain Rule

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In this section we compute partial derivatives using the chain rule.

For differentiable functions of one variable, the chain rule states that if $y = f(g(x))$ , then $y' = \frac {d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$ To extend this to the multi-variable setting, we will need to express this rule in Leibniz notation. If instead of writing $y = f(g(x))$ we write that $y = f(u)$ where $u = g(x)$ . Then the chain rule becomes $\frac {dy}{dx} = \frac {dy}{du} \cdot \frac {du}{dx}$

Now, we move to the multi-variable case. If $z = f(x,y)$ where $x$ and $y$ are functions of $t$ , then ultimately, $z$ is a function of $t$ . A change in $t$ causes a change in both $x$ and $y$ which in turn causes a change in $z$ leading to the following version of the chain rule: $\frac {dz}{dt} = \frac {\partial z}{\partial x} \cdot \frac {dx}{dt} + \frac {\partial z}{\partial y} \cdot \frac {dy}{dt}$

Example 1 Let $z = x^2 + y^2 + 3xy$ and let $x = 4\cos 5t$ and $y = 4\sin 5t$ . Find $\frac {dz}{dt}$ .
From the chain rule, we have \begin{align*} \frac{dz}{dt} &= \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\\ &= \left (2x + 3y\right ) \cdot \left (-20 \sin 5t\right ) + \left (2y + 3x\right ) \cdot \left (20 \cos 5t\right )\\ &= -20(8 \cos 5t + 12 \sin 5t) \cdot \sin 5t + 20 (8 \sin 5t + 12 \cos 5t) \cdot \cos 5t \\ &= -160 \cos 5t \sin 5t -240 \sin ^2 5t + 160 \sin 5t \cos 5t + 240 \cos ^2 5t\\ &= 240 (\cos ^2 5t - \sin ^2 5t)\\ &= 240 \cos 10t \end{align*}

Note, that it may have been easier to find $\frac {dz}{dt}$ by writing $z$ as a function of $t$ at the outset: \begin{align*} z &= x^2 + y^2 + 3xy \\ &= 16 \cos ^2 5t + 16 \sin ^2 5t + 48 \cos 5t \sin 5t\\ & = 16 + 24 \sin 10t \end{align*}

Now, the ordinary (single variable) chain rule gives $\frac {dz}{dt} = 240 \cos 10t$ which agrees with our previous answer.

(Problem 1a) Let $z = x^2 + y^2 + 4xy$ and let $x = \cos t$ and $y = \sin t$ . Use the chain rule to find $\frac {dz}{dt}$

(Problem 1b) Let $z = x^3 y^3 + 3x - 3y$ and let $x = 2\cos \pi t$ and $y = 2\sin \pi t$ . Find $\frac {dz}{dt}\bigg |_{t = 0}$

We next look at an extension of the previous chain rule to the case where $x$ and $y$ are themselves functions of two variables. Let $z$ be a function of $x$ and $y$ and let $x$ and $y$ both be functions of $s$ and $t$ . Then $z$ is ultimately a function of $s$ and $t$ and the chain rule says $\frac {\partial z}{\partial s} = \frac {\partial z}{\partial x} \cdot \frac {\partial x}{\partial s} + \frac {\partial z}{\partial y} \cdot \frac {\partial y}{\partial s}$

and

$\frac {\partial z}{\partial t} = \frac {\partial z}{\partial x} \cdot \frac {\partial x}{\partial t} + \frac {\partial z}{\partial y} \cdot \frac {\partial y}{\partial t}$

Example 2 Let $z = x^3 + xy^4, x = se^t$ and $y = \ln (st)$ . Find $\frac {\partial z}{\partial s}\bigg |_{s = 1, t = 2}$ and $\frac {\partial z}{\partial t}\bigg |_{s = 1, t = 2}$ .

We have: \begin{align*} \frac{\partial z}{\partial s} &= \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}\\ &= (3x^2 + y^4)e^t + 4xy^3\cdot \frac{1}{st} \cdot t\\ &= (3x^2 + y^4)e^t + \frac{4xy^3}{s} \end{align*}

When $s = 1$ and $t = 2$ , we have $x = e^2$ and $y = \ln (2)$ , hence $\frac {\partial z}{\partial s}\bigg |_{s = 1, t = 2} = (3e^4 + \ln ^4(2))e^2 + 4e^2 \ln ^3(2)$

As for the other partial derivative, we have: \begin{align*} \frac{\partial z}{\partial t} &= \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t}\\ &= (3x^2 + y^4)(se^t) + 4xy^3\cdot \frac{1}{st} \cdot s\\ &= se^t(3x^2 + y^4) + \frac{4xy^3}{t} \end{align*}

and $\frac {\partial z}{\partial t}\bigg |_{s = 1, t = 2} = e^2(3e^4 + \ln ^4(2)) + 2e^2 \ln ^3(2)$

(Problem 2) Let $z = e^x \sin y$ where $x = s^2 - t^2 - s - t$ and $y = \pi st$ . Find $\frac {\partial z}{\partial s}\bigg |_{s = 2, t = 1}$ and $\frac {\partial z}{\partial t}\bigg |_{s = 2, t = 1}$

Here is a video solution of problem 2:

0.1 Polar Coordinates

The polar coordinates $(r, \theta )$ of a point in the $xy$ -plane give the distance from the origin and the angle from the positive $x$ -axis. The relationship to standard rectangular coordinates $(x,y)$ is given by $x = r \cos \theta \quad \text {and} \quad y = r \sin \theta$

If $z = f(x,y)$ , then we can rewrite $z$ using polar coordinates as $z = f(r\cos \theta , r \sin \theta )$ and we can use the chain rule to compute the partial derivatives of $z$ with respect to $r$ and $\theta$ : \begin{align*} \frac{\partial z}{\partial r} &= \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial r}\\ &= \cos \theta \frac{\partial z}{\partial x} + \sin \theta \frac{\partial z}{\partial y} \end{align*}

and \begin{align*} \frac{\partial z}{\partial \theta } &= \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial \theta } + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial \theta }\\ &= -r\sin \theta \frac{\partial z}{\partial x} + r \cos \theta \frac{\partial z}{\partial y} \end{align*}

Example 3 Let $z = \ln (x^2 + y^2)$ where $x = r\cos \theta$ and $y = r\sin \theta$ . Find $\frac {\partial z}{\partial r}$ and $\frac {\partial z}{\partial \theta }$ .
Working directly with the chain rule, we have $\frac {\partial z}{\partial r} = \frac {2x}{x^2+y^2} \cos \theta + \frac {2y}{x^2+y^2} \sin \theta$ and $\frac {\partial z}{\partial \theta } = \frac {2x}{x^2+y^2} (-r\sin \theta ) + \frac {2y}{x^2+y^2} (r\cos \theta )$ Using the relations $x = r\cos \theta$ and $y = r\sin \theta$ , we can simplify each of these derivatives. However, we can also use these relations before applying the chain rule to write $z$ in terms of $r$ and $\theta$ . Noting that $x^2 + y^2 = r^2 \cos ^2 \theta + r^2 \sin ^2 \theta = r^2$ we have $z = \ln (r^2) = 2\ln r$ and hence $\frac {\partial z}{\partial r} = \frac {2}{r} \quad \text {and} \quad \frac {\partial z}{\partial \theta } = 0$

(Problem 3a) Let $z = e^{-(x^2 + y^2)}$ where $x = r\cos \theta$ and $y = r\sin \theta$ . Find $\frac {\partial z}{\partial r}$ and $\frac {\partial z}{\partial \theta }$ using the chain rule, and then again, by first writing $z$ as a function of $r$ and $\theta$ directly. Compare your answers.

(Problem 3b) Let $z = \tan ^{-1}\left (\frac {y}{x}\right )$ where $x = r\cos \theta$ and $y = r\sin \theta$ . Find $\frac {\partial z}{\partial r}$ and $\frac {\partial z}{\partial \theta }$ using the chain rule, and then again, by first writing $z$ as a function of $r$ and $\theta$ directly. Compare your answers.

Here is a video solution of problem 3b:

Example 4 If $z = f(x,y)$ where $x = r\cos \theta$ and $y = r \sin \theta$ , find $z_{rr}$ .
We have already established that $z_r = z_x \cos \theta + z_y \sin \theta$ where $z_x$ and $z_y$ are each functions of both $x$ and $y$ . By the chain rule, we have \begin{align*} z_{rr} &= \cos \theta \frac{\partial }{\partial r} z_x + \sin \theta \frac{\partial }{\partial r} z_y\\ &= \cos \theta \left [z_{xx}\cos \theta + z_{xy} \sin \theta \right ] + \sin \theta \left [ z_{yx} \cos \theta + z_{yy} \sin \theta \right ]\\ &= z_{xx}\cos ^2 \theta + 2z_{xy}\cos \theta \sin \theta + z_{yy}\sin ^2 \theta \end{align*}

(Problem 4) If $z = f(x,y)$ where $x = r\cos \theta$ and $y = r \sin \theta$ , find $z_{\theta \theta }$ .

Here is a video solution of problem 4:

0.2 General Chain Rule

Suppose $z = f(x_1, x_2, ..., x_n)$ and each of the variables $x_i$ for $i = 1, 2, ..., n$ is a function of the variables $t_1, t_2,..., t_m$ . Then for $1 \leq j \leq m$ , \begin{align*} \frac{\partial z}{\partial t_j} &= \frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_j} + \frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_j} + \cdots + \frac{\partial z}{\partial x_n}\frac{\partial x_n}{\partial t_j}\\ &= \sum _{i = 1}^n \frac{\partial z}{\partial x_i}\frac{\partial x_i}{\partial t_j} \end{align*}

Example 5 Suppose $w = f(x, y, z)$ where $x, y$ and $z$ are functions of $r, s$ and $t$ . Find an expression for $\frac {\partial w}{\partial t}$ .
Using the general form of the chain rule, we have $\frac {\partial w}{\partial t}= \frac {\partial w}{\partial x}\frac {\partial x}{\partial t}+\frac {\partial w}{\partial y}\frac {\partial y}{\partial t} +\frac {\partial w}{\partial z}\frac {\partial z}{\partial t}$

(Problem 5a) Suppose $w = f(x, y, z)$ where $x, y$ and $z$ are functions of $r, s$ and $t$ . Find an expression for $\frac {\partial w}{\partial s}$ .

(Problem 5b) Suppose $v = f(w, x, y, z)$ where $w, x, y$ and $z$ are functions of $s$ and $t$ . Find an expression for $\frac {\partial v}{\partial t}$ .

2025-05-26 16:55:31