2.4 The Chain Rule

Review §2.4.1 in the CLP-3 text.

(a) $\pdiff{h}{x}(x,y) =\pdiff{f}{x}\big(x,u(x,y)\big) +\pdiff{f}{u}\big(x,u(x,y)\big) \pdiff{u}{x}(x,y)$

(b) $\diff{h}{x}(x) =\pdiff{f}{x}\big(x,u(x),v(x)\big) +\pdiff{f}{u}\big(x,u(x),v(x)\big) \diff{u}{x}(x) +\pdiff{f}{v}\big(x,u(x),v(x)\big) \diff{v}{x}(x)$

(c) $\pdiff{h}{x}(x,y,z) =\pdiff{f}{u}\big(u(x,y,z),v(x,y),w(x)\big) \pdiff{u}{x}(x,y,z) +\pdiff{f}{v}\big(u(x,y,z),v(x,y),w(x)\big) \pdiff{v}{x}(x,y)$ $+\pdiff{f}{w}\big(u(x,y,z),v(x,y),w(x)\big) \diff{w}{x}(x)$

(c) We'll start with part (c) and follow the procedure given in §2.4.1 in the CLP-3 text. We are to compute the derivative of $h(x,y,z)=f\big(u(x,y,z),v(x,y),w(x)\big)$ with respect to $x$. For this function, the template of Step 2 in §2.4.1 is

Note that

Now we fill in the blanks with every variable that makes sense. In particular, since $f$ is a function of $u$, $v$ and $w$, it may only be differentiated with respect to $u$, $v$ and $w$. So we add together three copies of our template — one for each of $u$, $v$ and $w$:

Since $w$ is a function of only one variable, we use the ordinary derivative symbol $\diff{w}{x}$, rather than the partial derivative symbol $\pdiff{w}{x}$ in the third copy. Finally we put in the only functional dependence that makes sense. The left hand side is a function of $x$, $y$ and $z$, because $h$ is a function of $x$, $y$ and $z$. Hence the right hand side must also be a function of $x$, $y$ and $z$. As $f$ is a function of $u$, $v$ and $w$, this is achieved by evaluating $f$ at $u=u(x,y,z)$, $v=v(x,y)$ and $w=w(x)$.

(a) We again follow the procedure given in §2.4.1 in the CLP-3 text. We are to compute the derivative of $h(x,y)=f\big(x,u(x,y)\big)$ with respect to $x$. For this function, the template of Step 2 in §2.4.1 is

Now we fill in the blanks with every variable that makes sense. In particular, since $f$ is a function of $x$ and $u$, it may only be differentiated with respect to $x$, and $u$. So we add together two copies of our template — one for $x$ and one for $u$:

In $\diff{x}{x}$ we are to differentiate the (explicit) function $x$ (i.e. the function $F(x)=x$) with respect to $x$. The answer is of course $1$. So

Finally we put in the only functional depedence that makes sense. The left hand side is a function of $x$, and $y$, because $h$ is a function of $x$ and $y$. Hence the right hand side must also be a function of $x$ and $y$. As $f$ is a function of $x$, $u$, this is achieved by evaluating $f$ at $u=u(x,y)$.

(b) Yet again we follow the procedure given in §2.4.1 in the CLP-3 text. We are to compute the derivative of $h(x)=f\big(x,u(x),v(x)\big)$ with respect to $x$. For this function, the template of Step 2 in §2.4.1 is

(As $h$ is function of only one variable, we use the ordinary derivative symbol $\diff{h}{x}$ on the left hand side.) Now we fill in the blanks with every variable that makes sense. In particular, since $f$ is a function of $x$, $u$ and $v$, it may only be differentiated with respect to $x$, $u$ and $v$. So we add together three copies of our template — one for each of $x$, $u$ and $v$:

Finally we put in the only functional depedence that makes sense.

This is a visualization, in a simplified setting, of Example 2.4.10 in CLP3.

To visualize, in a simplified setting, the situation from Example 2.4.10 in CLP3, note that $w'(x)$ is the rate of change of $z$ as we slide along the blue line, while $f_x(x,y)$ is the change of $z$ as we slide along the orange line.

In the approximation $f_x(x,y)\approx \frac{\Delta f}{\Delta x}$, starting at the point $P_0$, $\Delta x=x_2-x_1$ and $\Delta f=z_2-z_1$.

In the approximation $\diff{w}{x}\approx \frac{\Delta w}{\Delta x}$, starting at the point $P_0$, $\Delta x=x_2-x_1$ again, and $\Delta w=z_1-z_1=0$.

To visualize, in a simplified setting, the situation from Example 2.4.10 in CLP3, note that $w'(x)$ is the rate of change of $z$ as we slide along the blue line, while $f_x(x,y)$ is the change of $z$ as we slide along the orange line.

In the partial derivative $f_x(x,y)\approx \frac{\Delta f}{\Delta x}$, we let $x$ change, while $y$ stays the same. Necessarily, that forces $f$ to change as well. Starting at point $P_0$, if we move $x$ but keep $y$ fixed, we end up at $P_2$. According to the labels on the diagram, $\Delta x$ is $x_2-x_1$, and $\Delta f$ is $z_2-z_1$.

The function $w(x)$ is a constant function, so we expect $w'(x)=0$. In the approximation $\diff{w}{x}\approx \frac{\Delta w}{\Delta x}$, we let $x$ change, but $w$ stays the same. Necessarily, to stay on the surface, this forces $y$ to change. Starting at point $P_0$, if we move $x$ but keep $z=f(x,y)$ fixed, we end up at $P_1$. According to the labels on the diagram, $\Delta x$ is $x_2-x_1$ again, and $\Delta w=z_1-z_1=0$.

To compare the two situations, note the first case has $\Delta y =0$ while the second case has $\Delta f =0$.

Pay attention to which variables change, and which are held fixed, in each context.

$\diff{w}{t}=1$ and $f_t=5$. $f_t$ gives the rate of change of $f(x,y,t)$ as $t$ varies while $x$ and $y$ are held fixed. $\diff{w}{t}$ gives the rate of change of $f\big(x(t),y(t),t\big)$. For the latter all of $x=x(t)$, $y=y(t)$ and $t$ are changing at once.

We are told in the statement of the question that $w(t)= f\big(x(t),y(t),t\big)$. Applying the chain rule to $w(t)= f\big(x(t),y(t),t\big),$ by following the procedure given in §2.4.1 in the CLP-3 tex, gives

Substituting in the values given in the question

On the other hand, we are told explicitly in the question that $f_t$ is $5$. The reason that $f_t$ and $\diff{w}{t}$ are different is that

The basic assumption is that the three quantites $x$, $y$ and $z$ are not independent. Given any two of them, the third is uniquely determined. They are assumed to satisfy a relationship $F(x,y,z)=0$, which can be solved to

For example, saying that $F(x,y,z)=0$ determines $x=f(y,z)$ means that

for all $y$ and $z$. The equation

really means

So use $(*)$ to compute $\pdiff{f}{z}$. Use other equations similar to $(*)$ to compute $\pdiff{g}{x}$ and $\pdiff{h}{y}$.

See the solution.

The basic assumption is that the three quantites $x$, $y$ and $z$ are not independent. Given any two of them, the third is uniquely determined. They are assumed to satisfy a relationship $F(x,y,z)=0$, which can be solved to

As an example, if $F(x,y,z) = xyz-1$, then

In general, saying that $F(x,y,z)=0$ determines $x=f(y,z)$ means that

for all $y$ and $z$. Set $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$. Applying the chain rule to $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)$ (with $y$ and $z$ independent variables) gives

The equation $(*)$ says that $\mathcal{F}(y,z)=F\big(f(y,z),y,z\big)=0$ for all $y$ and $z$. So differentiating the equation $(*)$ with respect to $z$ gives

for all $y$ and $z$. Similarly, differentiating $F\big(x,g(x,z),z\big)=0$ with respect to $x$ and $F\big(x,y,h(x,y)\big)=0$ with respect to $y$ gives

If $(x,y,z)$ is any point satisfying $F(x,y,z)=0$ (so that $x=f(y,z)$ and $y=g(x,z)$ and $z=h(x,y)$), then

and

Is the $\pdiff{w}{x}$ on the left hand side really the same as the $\pdiff{w}{x}$ on the right hand side?

The problem is that $\pdiff{w}{x}$ is used to represent two completely different functions in the same equation. See the solution for more details.

The problem is that $\pdiff{w}{x}$ is used to represent two completely different functions in the same equation. The careful way to write the equation is the following. Let $f(x,y,z)$ and $g(x,y)$ be continuously differentiable functions and define $w(x,y)=f\big(x,y,g(x,y)\big)$. By the chain rule,

While $w(x,y)=f\big(x,y,g(x,y)\big)$, it is not true that $\pdiff{w}{x}(x,y) =\pdiff{f}{x}\big(x,y,g(x,y)\big)$. For example, take $f(x,y,z)=x-z$ and $g(x,y)=x$. Then $w(x,y)=f\big(x,y,g(x,y)\big)=x-g(x,y)=0$ for all $(x,y)$, so that $\pdiff{w}{x}(x,y)=0$ while $\pdiff{f}{x}(x,y,z)=1$ for all $(x,y,z)$.

To avoid the chain rule, write $w$ explicitly as a function of $s$ and $t$.

$w_s(s,t)=2s(t^2+1)$ $w_t(s,t)=s^2(2t)$

Method 1: Since $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ with $x(s,t)=st$, $y(s,t)=s\cos t$ and $z(s,t)=s\sin t$ we can write out $w(s,t)$ explicitly:

• Method 2:* The question specifies that $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ with $x(s,t)=st$, $y(s,t)=s\cos t$ and $z(s,t)=s\sin t$. That is, $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ with $W(x,y,z)= x^2+y^2+z^2$. Applying the chain rule to $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ and noting that $\pdiff{W}{x}=2x$, $\pdiff{W}{y}=2y$, $\pdiff{W}{z}=2z$, gives

Start by setting $F(x,y)=f(2x+3y,xy)$. It might also help to define $g(x,y)=2x+3y$ and $h(x,y)=xy$.

We have

All functions on the right hand side have arguments $(2x+3y,xy)$. The notation $f_{21}$, for example, means first differentiate with respect to the second argument and then differentiate with respect to the first argument.

By definition,

We'll compute the derivatives from the inside out. Let's call $F(x,y)=f(2x+3y,xy)$ so that the innermost derivative is $G(x,y)=\pdiff{}{y}f(2x+3y,xy)=\pdiff{}{y}F(x,y)$. By the chain rule

Here the subscript $1$ means take the partial derivative of $f$ with respect to the first argument while holding the second argument fixed, and the subscript $2$ means take the partial derivative of $f$ with respect to the second argument while holding the first argument fixed. Next call the middle derivative $H(x,y)=\pdiff{}{y}\left(\pdiff{}{y}f(2x+3y,xy)\right)$ so that

By the chain rule (twice),