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

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 $\Delta w=z_1-z_1=0$.

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.

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.

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.

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

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

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.

Here $f_1$ denotes the partial derivative of $f$ with respect to its first argument, $f_{12}$ is the result of first taking one partial derivative of $f$ with respect to its first argument and then taking a partial derivative with respect to its second argument, and so on.

Start by showing that, because $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$, the second derivative $\frac{\partial^2g}{\partial s^2} = 2\frac{\partial^2f}{\partial x \partial y}$.

See the solutions.

The notation in the statement of this question is horrendous — the symbol $z$ is used with two different meanings in one equation. On the left hand side, it is a function of $x$ and $y$, and on the right hand side, it is a function of $s$ and $t$.
Unfortunately that abuse of notation is also very common. Until you get used to it, undo this notation conflict by renaming the function of $s$ and $t$ to $F(s,t)$. That is, $F(s,t) = f\big(2s+t\,,\,s-t\big)$.

Then, evaluate each term on the right-hand side of the equation.

$a=5$ and $b=c=2$.

Let $u(x,y) = x^2 - y^2$ , and $v(x,y) = 2xy$. Then $F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big)$.

(b) Since $F$ is a function of only one variable, the chain rule for (say) $\frac{\partial}{\partial x} F\big(xe^{-y^2}\big)$ has only one term.

(a) $\pdiff{u}{x}(x,y) = \frac{e^y}{x}$, $\pdiff{u}{y}(x,y) = e^y\,\ln (x) -y^2\,e^y -2y e^y$

(b) See the solution.

At some point, you'll be using the chain rule you learned in first-semester calculus.

$-54$

Just compute the first order partial derivatives of $w(x,y,z)$.

See the solution.

The function $\pdiff{f}{x}$ depends on both $x$ and $y$, so don't forget to account for both of these when you take its partial derivative.

(a)

(b)

with all of the partial derivatives of $f$ evaluated at $(r\cos t\,,\,r\sin t)$.

$28$

Just compute $\frac{\partial G}{\partial t}$, $\frac{\partial^2 G}{\partial \gamma^2}$ and $\frac{\partial^2 G}{\partial s^2}$.

$A=2$.

$10a$

Use implicit differentiation to find $\pdiff{z}{x}(x,y)$ and $\pdiff{z}{y}(x,y)$.

See the solution.

(a) $w_{ss}(s,t) =4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t) +9\,u_{yy}(2s + 3t, 3s - 2t)$

(b) $A=-1$

This question uses bad (but standard) notation, in that the one symbol $f$ is used for two different functions, namely $f(x,y)$ and $f(r,\theta)=f(x,y)\big|_{x=r\cos\theta,\,y=r\sin\theta}$. Until you get used to it, undo this notation conflict by renaming the function of $r$ and $\theta$ to $F(r,\theta)$. That is, $F(r,\theta) = f\big(r\cos\theta\,,\,r\sin\theta\big)$. Similarly, rename $g$, viewed as a function of $r$ and $\theta$, to $G(r,\theta)$. That is, $G(r,\theta) = g\big(r\cos\theta\,,\,r\sin\theta\big)$.

(a)

with the arguments of $f_x$, $f_y$, $f_{xx}$, $f_{xy}$ and $f_{yy}$ all being $\big(r\cos\theta\,,\,r\sin\theta\big)$.

(b)

$\vnabla z(1,2)=\llt -47,108\rgt$

(b) Think of $x=u^3-3uv^2$, $y=3u^2v-v^3$ as two equations in the two unknowns $u$, $v$ with $x$, $y$ just being given parameters. The question implicitly tells us that those two equations can be solved for $u$, $v$ in terms of $x$, $y$, at least near $(u,v)=(2,1)$, $(x,y)=(2,11)$. That is, the question implicitly tells us that the functions $u(x,y)$ and $v(x,y)$ are determined by $x=u(x,y)^3-3u(x,y)\,v(x,y)^2$, $y=3u(x,y)^2v(x,y)-v(x,y)^3$. Then $z(x,y)$ is determined by $z(x,y)=u(x,y)^2-v(x,y)^2$.

(a) See the solution. (b) $\frac{4}{15}$

The question tells us that $x(u,v)$ and $y=y(u,v)$ ar eimplicitly determined by