4.5 The Chain Rule

← Chapter 4 §4.5 The Chain Rule 链式法则 · Chapter 4 · Differentiation of Functions of Several Variables

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Access for free at openstax.org · CC BY-NC-SA 4.0

Objective	Takeaways; say them first, then click
4.5.1 State the chain rules for one or two independent variables.	$dz/dt=z_x\,x'+z_y\,y'$ ; $\partial z/\partial u=z_x\,x_u+z_y\,y_u$ , a partial on the left
4.5.2 Use tree diagrams as an aid to understanding the chain rule.	multiply along each branch ending in the variable, add the branches; one term per intermediate variable
4.5.3 Perform implicit differentiation of a function of two or more variables.	$dy/dx=-f_x/f_y$ ; $\partial z/\partial x=-f_x/f_z$ ; tangent line at a point

Mistake	Result	Fix
Only one route	$8x\cos t$ in Example 4.26a	one term per branch: add $(6y)(-\sin t)$
Answer left in $x$ and $y$	$8x\cos t-6y\sin t$ "in terms of $t$ "	substitute $x(t)$ , $y(t)$ last
$dz/du$ with two independent variables	hides that $v$ is held fixed	$\partial z/\partial u$
Partials evaluated at $t$	$f_x(t,t)$ instead of $f_x(x(t),y(t))$	partials at the point on the path
Dropping the minus in $-f_x/f_y$	slope $-\tfrac74$ at $(2,1)$ in Example 4.30a	$f$ is constant: the two routes cancel
$f_y/f_x$ upside down	slope $\tfrac47$ instead of $\tfrac74$	the variable you differentiate by is on top
Misreading the tree	a $v$ -branch in $\partial z/\partial u$	keep only leaves labelled with your variable

#	task	answer
4.23	$z=x^2-3xy+2y^2$ , $x=3\sin2t$ , $y=4\cos2t$	$\dfrac{dz}{dt}=-46\sin4t-72\cos4t$
4.24	$z=\dfrac{2x-y}{x+3y}$ , $x=e^{2u}\cos3v$ , $y=e^{2u}\sin3v$	$\dfrac{\partial z}{\partial u}=0$ , $\ \dfrac{\partial z}{\partial v}=\dfrac{-21}{(3\sin3v+\cos3v)^2}$
4.25	$w=\dfrac{x+2y-4z}{2x-y+3z}$ , $x=e^{2u}\cos3v$ , $y=e^{2u}\sin3v$ , $z=e^{2u}$	$\dfrac{\partial w}{\partial u}=0$ , $\ \dfrac{\partial w}{\partial v}=\dfrac{15-33\sin3v+6\cos3v}{(3+2\cos3v-\sin3v)^2}$
4.26	tree for $w=f(x,y)$ , $x,y$ of $t,u,v$	6 leaves; e.g. $\dfrac{\partial w}{\partial t}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial t}$ (tree lab)
4.27	$x^2+xy-y^2+7x-3y-26=0$ at $(3,-2)$	$\dfrac{dy}{dx}=-\dfrac{11}{4}$ ; $\ y=-\dfrac{11}{4}x+\dfrac{25}{4}$

#	task	answer	key step
215	$w=xy\cos z$ , $x=t$ , $y=t^2$ , $z=\arcsin t$	$\dfrac{3t^2-4t^4}{\sqrt{1-t^2}}$	$\cos(\arcsin t)=\sqrt{1-t^2}$
217	$w=5x^2+2y^2$ , $x=-3s+t$ , $y=s-4t$	$94s-46t$ ; $\ -46s+74t$	two branches each
219	$f=xy$ , polar $x,y$ : $\partial f/\partial r$	$r\sin2\theta$	$2r\sin\theta\cos\theta$
221	$f=x^2+y^2$ , $x=t$ , $y=t^2$	$2t+4t^3$	both methods agree
223	$f=xy$ , $x=1-\sqrt t$ , $y=1+\sqrt t$	$-1$	$f=1-t$
227	$w=x^2+y^2+z^2$ , $x=\cos t$ , $y=\sin t$ , $z=e^t$	$2e^{2t}$	$w=1+e^{2t}$
229	$u=e^x\sin y$ , $x=-\ln2t$ , $y=\pi t$ at $x=\ln2$	$\sqrt2(\pi-4)$	$t=\tfrac14$
231	$x^3+y^2x-3=0$	$-\dfrac{3x^2+y^2}{2xy}$	$-f_x/f_y$
237	$e^{xy}+ye^y=1$	$-\dfrac{ye^{xy}}{xe^{xy}+e^y(1+y)}$	product rule in $f_y$
243	$z=x/y$ , $x=2\cos u$ , $y=3\sin v$	$\dfrac{-2\sin u}{3\sin v}$ ; $\ \dfrac{-2\cos u\cos v}{3\sin^2v}$	one branch each
245	$z=xye^{x/y}$ , polar, $r=2$ , $\theta=\pi/6$	$\sqrt3\,e^{\sqrt3}$ ; $\ (2-4\sqrt3)e^{\sqrt3}$	$(x,y)=(\sqrt3,1)$
251	cylinder $V=\pi x^2y$ , $x=t/2$ , $y=t/3$ at $x=2$	$4\pi$	$t=4$
253	cone: $r'=3$ , $h'=-2$ at $r=13$ , $h=18$ (p373)	$\tfrac{1066\pi}{3}\approx1116.31$ cm³/min	$468\pi-\tfrac{338\pi}{3}$
255	box surface area, each side $+0.5$ in/min	12 in²/min	$2(x+y+z)$
257	the fly	2 °C/s	$4\cdot\tfrac14+3\cdot\tfrac13$

The Chain Rule

Moving along a path, the height changes through $x$ and through $y$ : add the two rates

The routes add because, up close, a differentiable surface is its tangent plane

The chain rule is a dot product: $\langle f_x,f_y\rangle\cdot\langle x',y'\rangle$

Walking the unit circle under $z=4x^2+3y^2$ , you climb at $2\sin t\cos t$

Four derivatives, two products, then write the answer in $t$

Along a path, the two route terms add up to the slope of $z(t)$

With $x=e^{2t},\ y=e^{-t}$ : $\ \frac{d}{dt}\sqrt{x^2-y^2}=\frac{2e^{6t}+1}{e^t\sqrt{e^{6t}-1}}$

A fly on a hot plate warms at $4\cdot\tfrac14+3\cdot\tfrac13=2$ °C/s

Wider beats shorter: the cone gains $\tfrac{1066\pi}{3}$ cm³/min

A tree diagram: multiply along each branch, add the branches

Two independent variables: follow only the branches that end in your variable

Six partials, two sums: $\frac{\partial z}{\partial u}=38u+18v$ , $\ \frac{\partial z}{\partial v}=18u+34v$

$\partial z/\partial u=0$ : changing $u$ only slides the point along a ray where $z$ is constant

Scaling both inputs is a path too: the chain rule gives $x\,f_x+y\,f_y=n\,f$

Any number of variables: one term for every intermediate variable

Nine partials give $\frac{\partial w}{\partial u}=2e^{2u}(3\sin^2v-2\sin v\cos v+4)$

Three intermediate × three independent: nine leaves, three formulas of three terms

Pick a variable: the branches that end in it are the terms of its formula

On the curve $f(x,y)=0$ , the slope is $-f_x/f_y$ — no solving for $y$

Implicit differentiation is the chain rule applied to $f(x,y(x))=0$

At $(2,1)$ the tilted ellipse has slope $\tfrac74$ : tangent $y=\tfrac74x-\tfrac52$

For a surface $f(x,y,z)=0$ , divide by $-f_z$

Along $f(x,y)=0$ the tangent slope is $-f_x/f_y$ wherever $f_y\neq0$

The tangent at $(3,-2)$ is $y=-\frac{11}{4}x+\frac{25}{4}$

Count the routes, multiply along each, add them up

Common mistakes

Practice and answers

The Chain Rule

Moving along a path, the height changes through xxx and through yyy: add the two rates

The routes add because, up close, a differentiable surface is its tangent plane

The chain rule is a dot product: ⟨fx,fy⟩⋅⟨x′,y′⟩\langle f_x,f_y\rangle\cdot\langle x',y'\rangle⟨fx​,fy​⟩⋅⟨x′,y′⟩

Walking the unit circle under z=4x2+3y2z=4x^2+3y^2z=4x2+3y2, you climb at 2sin⁡tcos⁡t2\sin t\cos t2sintcost

Four derivatives, two products, then write the answer in ttt

Along a path, the two route terms add up to the slope of z(t)z(t)z(t)

With x=e2t, y=e−tx=e^{2t},\ y=e^{-t}x=e2t, y=e−t: ddtx2−y2=2e6t+1ete6t−1\ \frac{d}{dt}\sqrt{x^2-y^2}=\frac{2e^{6t}+1}{e^t\sqrt{e^{6t}-1}} dtd​x2−y2​=ete6t−1​2e6t+1​

A fly on a hot plate warms at 4⋅14+3⋅13=24\cdot\tfrac14+3\cdot\tfrac13=24⋅41​+3⋅31​=2 °C/s

Wider beats shorter: the cone gains 1066π3\tfrac{1066\pi}{3}31066π​ cm³/min