6 concepts17 worked examples27 exercises3 exam-level5 figures
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09Tangent planes, linear approximation, and the chain rule for functions of several variables
A weather balloon carries a thermometer through the air. The temperature at each point of the sky is a fixed number $T$ that depends on where you are, on longitude $x$ and altitude $y$. The balloon is at one spot now; a second later it has drifted a little sideways and risen a little. The reading on the thermometer changes. Nobody handed you a formula for the temperature along the balloon's path; you were only given $T$ as a function of position and the two speeds of the balloon. How fast is the number on the dial moving?
By the end you can turn any surface into the flat plane that touches it, use that plane to estimate values a calculator would need, and compute the rate of a quantity that depends on several moving inputs, whether one clock or several dials drive the motion.
In 60 seconds
Near a point, a differentiable surface and its tangent plane are interchangeable; and when a quantity is built from inputs that themselves move, its rate is the sum, over every intermediate variable, of one times the rate of that intermediate.
Tangent plane / at (a,b)
$$L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$
you need the flat approximation to $f$, or an estimate of $f$ at an awkward point near a friendly one
Total differential
$$dz=f_x\,dx+f_y\,dy$$
small changes $dx,dy$ are given and you want the resulting change or error in $z$
$z$ is defined implicitly by $F(x,y,z)=0$ and $F_z\ne 0$
Three most common mistakes
Writing the tangent plane as $f_x(a,b)(x-a)+f_y(a,b)(y-b)$ and dropping the height $f(a,b)$, so the plane passes through the origin instead of touching the surface.
Using $\partial z/\partial t$ when $x,y$ depend on $t$ alone: the answer is an ordinary derivative $dz/dt$, and forgetting a term because the tree had two branches.
In implicit differentiation, forgetting the minus sign in $-F_x/F_z$, or dividing by $F_y$ instead of $F_z$ when solving for $\partial z/\partial x$.
This section feeds the Quiz and Homework blocks (worth 10% each). Expect one routine tangent-plane-or-linearization item and one chain-rule item; the harder version mixes the two, or asks for a chain-rule identity in polar coordinates rather than a number.
How much time do you have?
10 minutes
You leave holding the two formulas the exam leans on hardest: the tangent plane / linearization, and the one-parameter chain rule. Both are on the card, both are drilled in B.
card, c-tangent-plane, c-chain-1, formula card
45 minutes
You can also read the tree diagram for several independent variables, take a total differential to bound a measurement error, and differentiate an equation implicitly for a slope.
card, c-tangent-plane, c-linear-approx, c-differentials, c-chain-1, c-chain-2, c-implicit, scaffolding ladder, practice C
full read
Everything above plus the definition of differentiability and the surface that has both partials yet no tangent plane, the polar-coordinate exam identity, and the mixed practice that forces you to pick the right tool without being told which section it came from.
everything in order, practice A, practice B, practice C, practice D, self check
By the end of this section
Write the equation of the tangent plane to a surface $z=f(x,y)$ at a given point.
Estimate a value of $f(x,y)$ with the linear approximation, and state when a function is differentiable.
Compute the total differential and use it to estimate the change in $z$ and to bound a measurement error.
Differentiate a composite $z=f(x,y)$ with $x,y$ functions of one variable $t$, obtaining $dz/dt$.
Apply the chain rule with a tree diagram when $x,y$ depend on several independent variables.
Differentiate implicitly to find $dy/dx$ from $F(x,y)=0$ and $\partial z/\partial x$ from $F(x,y,z)=0$.
Syllabus coverage
14.4
Tangent planes and linear approximations
Tangent plane, the linearization, the definition of differentiability, total differentials, and error estimation are the three opening concepts.
covered
14.5
The chain rule
One-parameter and several-parameter chain rule with tree diagrams, and implicit differentiation, are the last three concepts.
covered
Recall first
Partial derivatives
For $f(x,y)$, the partial $f_x$ is the ordinary derivative with respect to $x$ while $y$ is treated as a constant; $f_y$ treats $x$ as a constant. For $f(x,y)=x^2y^3$, $f_x=2xy^3$ and $f_y=3x^2y^2$.
Every formula in this section is built from $f_x$ and $f_y$; if you cannot produce them quickly the rest stalls.
Single-variable chain rule
$\dfrac{d}{dt}f(g(t))=f'(g(t))\,g'(t)$. For example $\dfrac{d}{dt}\sin(t^2)=\cos(t^2)\cdot 2t$.
The multivariable chain rule is this rule applied along each branch of a tree and then summed.
Single-variable linearization
The tangent line to $y=f(x)$ at $x=a$ is $L(x)=f(a)+f'(a)(x-a)$, used to approximate $f$ near $a$.
The tangent plane is the exact two-variable copy of this line, with one slope term for each input.
Equation of a plane
A plane through $(x_0,y_0,z_0)$ with normal vector $\langle A,B,C\rangle$ is $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$.
The tangent plane is one such plane; recognising the form lets you read its normal vector straight off.
Derivatives of the standard functions
$\dfrac{d}{dt}e^{t}=e^{t}$, and by the chain rule $\dfrac{d}{dt}e^{-t}=-e^{-t}$; $\dfrac{d}{dt}\sin t=\cos t$, $\dfrac{d}{dt}\cos t=-\sin t$.
Chain-rule problems here feed on exponentials and trig; a dropped minus sign is the commonest arithmetic slip.
A trace of a surface
Fixing $y=b$ in $z=f(x,y)$ leaves a curve $z=f(x,b)$ in a vertical plane; its slope at $x=a$ is exactly $f_x(a,b)$. Fixing $x=a$ gives the other trace, with slope $f_y(a,b)$.
The tangent plane is the unique plane whose two traces are the tangent lines of these two curves.
Try it yourself first (3 questions)
1§09.0 — a partial derivative, from last week●●○○○
Before tangent planes, make sure the two ingredients come out fast. This one has a built-in trap for anyone who slips back into one-variable habits.
Given
$f(x,y)=x^2y^3$
Find
(a) What is $f_x(x,y)$?
Hint 1/4
You want the rate as $x$ moves only, so $y$ is frozen into a constant.
Hint 2/4
Differentiate $x^2$ times the constant $y^3$; the constant rides along untouched.
derivative of the outer $\sin$ evaluated at the inner, times the inner's derivative
$$=\cos(t^2)\cdot 2t=2t\cos(t^2)$$
the inner derivative is $2t$
Answer $$\boxed{g'(t)=2t\cos(t^2)}$$
Check
At $t=0$ the answer is $0$; and $g(t)=\sin(t^2)$ is flat at $t=0$ because $t^2$ has a minimum there, so a zero slope is right.
3§09.0 — the equation of a plane●●○○○
The tangent plane is a plane; you should be able to write any plane from a point and a normal without thinking.
Given
the plane passes through $(1,2,3)$
its normal vector is $\langle 4,-1,2\rangle$
Find
(a) Which equation describes this plane?
Hint 1/4
A plane is fixed by one point on it and one direction perpendicular to it.
Hint 2/4
Use $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$ with $\langle A,B,C\rangle$ the normal.
Hint 3/4
Here $\langle A,B,C\rangle=\langle 4,-1,2\rangle$ and $(x_0,y_0,z_0)=(1,2,3)$.
Hint 4/4
So $4(x-1)-1(y-2)+2(z-3)=0$.
Show solutionSubstitute into the point-normal form
$$4(x-1)+(-1)(y-2)+2(z-3)=0$$
the normal supplies the coefficients; the point supplies what is subtracted
$$4x-y+2z=4-2+6=8$$
expand and collect the constants on the right
Answer $$\boxed{4x-y+2z=8}$$
Check
Test the point: $4(1)-2+2(3)=4-2+6=8$, so $(1,2,3)$ satisfies the equation.
Notation
symbol
reads as
means
watch out
$f_x,\ f_y$
the partial of f with respect to x, and with respect to y
the rate of $f$ when only $x$ moves (with $y$ held fixed), and when only $y$ moves; also written $\partial f/\partial x$ and $\partial f/\partial y$
$f_x(a,b)$ is a number; $f_x$ alone is a function. The plane and the differential use the number.
$L(x,y)$
the linearization of f at the base point
the tangent plane to $z=f(x,y)$ at $(a,b)$, read as a formula for approximate values of $f$ near $(a,b)$
$L$ carries its base point $(a,b)$ inside every coefficient; move the base point and all four numbers in $L$ change.
$dz$
the differential of z
$dz=f_x\,dx+f_y\,dy$, the change in $z$ predicted by the tangent plane over the input steps $dx,dy$
$dz$ is the estimate; $\Delta z=f(a+dx,b+dy)-f(a,b)$ is the truth. They are close, not equal.
$\partial z/\partial t\ \text{vs}\ dz/dt$
partial dee z by dee t, versus dee z by dee t
the partial holds every other independent variable fixed; the ordinary $d$ is used when $t$ is the only independent variable
Reaching for $\partial$ when there is a single independent variable is the most common notation error in this section.
Conventions used here
Partial versus ordinary d
When $z$ depends on inputs that all move with a single variable $t$, the result is an ordinary derivative and we write $dz/dt$. When the inputs depend on more than one independent variable, each result is a partial derivative and we write $\partial z/\partial s$. The symbol on the far left tells the reader how many dials are turning.
Where the partials are evaluated
In the tangent plane and the chain rule, $f_x$ and $f_y$ are evaluated at the base point or at the current point, never left as functions when a number is asked for. A partial derivative written without a point is a function; the plane and the rate want the number it takes at the point in question.
Differentiability is stronger than having partials
Saying a function is differentiable at a point means its graph has a genuine tangent plane there, so that the linear approximation error dies faster than the distance to the point. The mere existence of $f_x$ and $f_y$ does not guarantee this; a sufficient condition is that $f_x$ and $f_y$ are continuous near the point.
The differential symbols dx, dy
In a differential, $dx$ and $dy$ are the actual (small) changes you choose to make in the inputs, equal to $\Delta x$ and $\Delta y$. The output differential $dz$ is then only an estimate of the true change $\Delta z$; the two agree to first order and differ by an amount that shrinks faster than the step.
9.1The plane that touches a surface
The tangent plane packs the height and both partial slopes at a point into one flat equation.
We can already produce both partial derivatives of a surface $z=f(x,y)$. The one thing left is to assemble those two slopes into a single flat sheet that hugs the surface at a chosen point.
Solvable with what we have
Find the tangent line to a curve $y=f(x)$ at a point: one point, one slope $f'(a)$, done.
Compute $f_x$ and $f_y$ for a surface with $y$ or $x$ frozen.
Not solvable yet
Write down the flat object that touches a whole surface $z=f(x,y)$ at a point.
Use it to estimate the height of the surface a short step away.
A tangent line worked with one slope, so try $z=f(a,b)+f_x(a,b)(x-a)$ and call it the tangent to the surface.
Why it fails
That is only a line, sitting in the plane $y=b$. It follows the surface as $x$ moves but ignores everything that happens as $y$ moves. A surface is two-dimensional; the flat thing that matches it must be a plane, and a plane needs a $y$-slope too.
RuleTangent plane to z = f(x, y)
Conditions
$f$ has continuous partial derivatives near $(a,b)$ (so it is differentiable there)
Start at the height of the surface above the base point; then add the x-slope times how far you moved in x, and the y-slope times how far you moved in y.
The $\textcolor{#d1690a}{\text{tangent plane}}$ rests on the $\textcolor{#1f6feb}{\text{surface}}$ and meets it at the single point $P$. Move a little away from $P$ and the gap between plane and surface opens slowly; that slow opening is what makes the plane a good local stand-in for $f$.
Looks like this, but is not
The cone $z=\sqrt{x^2+y^2}$ looks like a smooth surface that should have a tangent plane everywhere, including at its tip $(0,0,0)$.
At the tip the surface comes to a sharp point, like the bottom of an ice-cream cone. No single flat plane can touch it there; indeed $f_x$ and $f_y$ do not even exist at $(0,0)$. A tangent plane needs the partials to exist first, and a smooth-looking picture is not a guarantee.
Tangent plane to z = x squared minus y squared at (2, 1)
Find the tangent plane to the surface $z=x^2-y^2$ at the point above $(2,1)$.
At $(3,4)$: $\tfrac{3}{5}(3)+\tfrac{4}{5}(4)=\tfrac{9+16}{5}=5$, matching $f(3,4)$. The constant genuinely vanished because the cone's tangent planes all pass through the origin.
Checkpoint
§09.1 — reading off a tangent plane●●○○○
The whole method is three numbers dropped into one formula. This checks that you keep the height and place the slopes on the right variable.
Given
$f(x,y)=xy$
base point $(2,3)$
Find
(a) Which is the tangent plane at the point above $(2,3)$?
Hint 1/4
You need three numbers: the height $f(2,3)$, and the two slopes $f_x(2,3)$ and $f_y(2,3)$.
Hint 2/4
For $f=xy$, $f_x=y$ and $f_y=x$, so at $(2,3)$ the slopes are $3$ and $2$; the height is $6$.
Hint 3/4
Drop them into $z=f(a,b)+f_x(x-a)+f_y(y-b)=6+3(x-2)+2(y-3)$.
$3(2)+2(3)-6=6=f(2,3)$; the plane meets the surface at the base point.
⚠ Dropping the height f(a,b)
the slope terms feel like the calculus, so the constant height reads as decoration
wrong$$z=f_x(a,b)(x-a)+f_y(a,b)(y-b)$$
right$$z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$
⚠ Putting each slope on the wrong variable
$f_x$ and $f_y$ get written down in a rush and attached to whichever bracket comes first
wrong$$z=f(a,b)+f_y(a,b)(x-a)+f_x(a,b)(y-b)$$
right$$z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$
9.2Using the plane as a calculator, and when you are allowed to
The tangent plane's height estimates f cheaply, and it is trustworthy exactly when f is differentiable.
Read the tangent plane as a rule for height instead of a geometric object, and it becomes a hand calculator: it trades a hard value of $f$ for one multiplication and one addition.
Near the base point, the surface and its tangent plane share almost the same height, so the plane's height is a good estimate of the function's.
Looks like this, but is not
Take $f(x,y)=\dfrac{xy}{x^2+y^2}$ with $f(0,0)=0$. Along each axis $f$ is constantly $0$, so both partials exist at the origin: $f_x(0,0)=f_y(0,0)=0$. It looks ready for a linear approximation.
But approach the origin along the line $y=x$: $f(x,x)=\dfrac{x^2}{2x^2}=\tfrac12$, not $0$. So $f$ is not even continuous at $(0,0)$, let alone differentiable, and the 'linear approximation' $L\equiv 0$ is wrong by $\tfrac12$ arbitrarily close in. Existence of both partials is not enough; that is why the theorem asks for the partials to be continuous.
step h
point
true f
estimate L
error
1.0
(4.0, 5.0)
6.4031
6.4000
0.0031
0.5
(3.5, 4.5)
5.7009
5.7000
0.0009
0.1
(3.1, 4.1)
5.14004
5.14000
0.00004
Shrink the step by a factor of 5 (from 0.5 to 0.1) and the error drops by about 20, close to $5^2=25$. The error of a tangent-plane estimate is second order in the step: halve the step and you roughly quarter the error. That is exactly why differentiability, defined below, is the right idea.
Estimating the length of the vector (3.02, 3.96)
Use a linear approximation to estimate $\sqrt{(3.02)^2+(3.96)^2}$ by hand.
Given
$f(x,y)=\sqrt{x^2+y^2}$
target point $(3.02,3.96)$
Find
a decimal estimate without a calculator
SolutionPick a friendly base point nearby
$$(a,b)=(3,4)$$
the closest point where $f$ and both partials are exact, and the steps $dx=0.02,\ dy=-0.04$ are small
Square the estimate: $4.980^2=24.8004$, while $(3.02)^2+(3.96)^2=9.1204+15.6816=24.802$. The two agree to four figures, so the estimate is sound.
A step of a few hundredths gives an error in the fourth decimal place, exactly the second-order behaviour the table predicted.
Where is f(x, y) = x squared plus x y differentiable?
State where $f(x,y)=x^2+xy$ is differentiable, and justify it with the theorem.
Given
$f(x,y)=x^2+xy$
Find
the set of points where a tangent plane exists
SolutionCheck the sufficient condition
$$f_x=2x+y,\qquad f_y=x$$
both partials, taken with the other variable frozen
$$2x+y\ \text{and}\ x\ \text{are polynomials, continuous on all of}\ \mathbb{R}^2$$
polynomials are continuous everywhere, which is exactly the hypothesis the theorem needs
Answer $$\boxed{f\ \text{is differentiable at every point}\ (a,b)}$$
Check
A quick sanity check: the counterexample failed because its partials were discontinuous at the origin; here nothing like that can happen, so a tangent plane exists everywhere.
Checkpoint
§09.2 — one step of a linear approximation●●○○○
You do not need the true value to answer; the estimate is pure arithmetic once you have the three numbers.
Given
$f(x,y)=\sqrt{x^2+y^2}$
base point $(3,4)$ with $f=5,\ f_x=0.6,\ f_y=0.8$
Find
(a) Estimate $f(3.1,3.9)$ using the linear approximation at $(3,4)$.
Hint 1/4
You are moving from $(3,4)$ to $(3.1,3.9)$, so find the two steps first.
Hint 2/4
The estimate is $L=f(a,b)+f_x\,dx+f_y\,dy$ with $dx=0.1$ and $dy=-0.1$.
Hint 3/4
Substitute: $L=5+0.6(0.1)+0.8(-0.1)$, using $f=5,\ f_x=0.6,\ f_y=0.8$.
Hint 4/4
So $L=5+0.06-0.08=4.98$.
Show solutionPlug into the plane
$$L=5+0.6(0.1)+0.8(-0.1)$$
the linear approximation with the two steps
$$=5+0.06-0.08=4.98$$
collect
Answer $$\boxed{L=4.98}$$
Check
True value $\sqrt{9.61+15.21}=\sqrt{24.82}=4.9820$; the estimate $4.98$ is right to three figures.
⚠ Using the wrong sign on a step
$dx$ and $dy$ are read as the target coordinates instead of the change from the base point
wrong$$L=5+0.6(3.1)+0.8(3.9)$$
right$$L=5+0.6(0.1)+0.8(-0.1)$$
⚠ Approximating where f is not differentiable
a formula exists, so it feels safe to linearize, even at a corner or a break
right$$\text{no tangent plane at}\ (0,0);\ \text{the tip is a corner}$$
9.3Small changes: the total differential
The total differential adds one partial-times-change term per input to estimate how much f moves.
Sometimes you do not want the value of $f$ at a new point but the change in $f$ when the inputs nudge a little. Rearranging the linear approximation gives that change its own name and its own symbol.
DefinitionTotal differential
Conditions
$f$ differentiable at the point
$dx=\Delta x,\ dy=\Delta y$ are the chosen input changes
$$\boxed{\,dz=f_x(x,y)\,dx+f_y(x,y)\,dy\,}$$
The predicted change in output is the x-slope times the change in x plus the y-slope times the change in y; each input contributes through its own partial.
For area $A=xy$, growing $x$ by $dx$ and $y$ by $dy$ adds two $\textcolor{#d1690a}{\text{strips}}$, of areas $y\,dx$ and $x\,dy$: their sum is exactly $dA$. The tiny $\textcolor{#2f9e44}{\text{corner}}$ $dx\,dy$ is what the differential leaves out, and it shrinks faster than either strip.
Looks like this, but is not
It is tempting to treat $dz$ as the exact change $\Delta z$, since for a straight line the tangent is the function.
For a genuinely curved surface they differ. In the area picture, $\Delta A=(x+dx)(y+dy)-xy=y\,dx+x\,dy+dx\,dy$, so $\Delta A-dA=dx\,dy$. The differential is missing the corner. It is small (a product of two small numbers) but it is not zero, which is why $dz$ is an estimate, not the truth.
How much does a heated metal plate grow?
A rectangular plate is $30$ cm by $24$ cm. Heating lengthens each side by $0.1$ cm. Use a differential to estimate the change in area, and compare with the exact change.
Given
$A=xy$ with $x=30,\ y=24$
$dx=dy=0.1$
Find
the estimated change in area $dA$, and the error in that estimate
SolutionWrite the differential of the area
$$dA=A_x\,dx+A_y\,dy=y\,dx+x\,dy$$
$A_x=y$ and $A_y=x$, since $A=xy$
$$dA=24(0.1)+30(0.1)=2.4+3=5.4$$
each strip: the long side times its own small growth
The missed piece $dx\,dy=0.01$ is a hundredth, while $dA=5.4$; the corner is smaller than $dA$ by a factor of hundreds, as a second-order term should be.
Whenever a quantity is a product, its differential is a sum of 'one factor times the other's change' terms, one per factor.
Worst-case error in a computed volume
The sides of a rectangular box are measured as $x=75$, $y=60$, $z=40$ cm, each with a possible error of at most $0.2$ cm. Estimate the largest resulting error in the computed volume $V=xyz$.
Given
$V=xyz$ at $(75,60,40)$
$|dx|,|dy|,|dz|\le 0.2$
Find
an estimate of the maximum error in $V$
SolutionDifferentiate the volume
$$dV=yz\,dx+xz\,dy+xy\,dz$$
the differential of a triple product: freeze two factors in each term
$$yz=2400,\quad xz=3000,\quad xy=4500$$
the three coefficients at the measured point
Push every term to its worst case
$$|dV|\le 2400(0.2)+3000(0.2)+4500(0.2)$$
take each error as large as allowed and all the same sign
$$=(9900)(0.2)=1980$$
factor out $0.2$ and add the coefficients
Answer $$\boxed{|dV|\lesssim 1980\ \text{cm}^3}$$
Check
As a relative error, $1980/180000=0.011=1.1\%$, which is roughly the sum of the three relative side errors $0.2/75+0.2/60+0.2/40$; a good order-of-magnitude check.
Three partials, but each is just a product of the two frozen sides.
Checkpoint
§09.3 — one differential●●○○○
Two terms, one per input. The only trap is forgetting a term or slipping in the corner.
Given
$z=x^2y$
at $(x,y)=(2,3)$
$dx=0.1,\ dy=0.1$
Find
(a) Estimate the change $dz$.
Hint 1/4
Write the differential in general first, then substitute the point and the steps.
Hint 2/4
For $z=x^2y$: $z_x=2xy$ and $z_y=x^2$, so $dz=2xy\,dx+x^2\,dy$.
Hint 3/4
At $(2,3)$: $z_x=2(2)(3)=12$ and $z_y=2^2=4$, with $dx=dy=0.1$.
Hint 4/4
So $dz=12(0.1)+4(0.1)=1.2+0.4=1.6$.
Show solutionPartials then substitute
$$dz=2xy\,dx+x^2\,dy$$
one term per input
$$=12(0.1)+4(0.1)=1.6$$
evaluate the partials at $(2,3)$, then the steps
Answer $$\boxed{dz=1.6}$$
Check
The exact change is $2.1^2(3.1)-2^2(3)=13.671-12=1.671$; the differential $1.6$ is close, low by the second-order part.
⚠ Including the corner term in dz
the exact change has a $dx\,dy$ piece, so it sneaks into the differential
wrong$$dz=f_x\,dx+f_y\,dy+dx\,dy$$
right$$dz=f_x\,dx+f_y\,dy$$
⚠ Losing a term in a product's differential
only the factor that changed 'a lot' gets a term; the other is dropped
wrong$$d(xy)=y\,dx$$
right$$d(xy)=y\,dx+x\,dy$$
9.4One clock driving everything: the chain rule in t
When one variable t drives x and y, dz/dt sums one partial-times-speed term for each of the two routes.
Back to the balloon. The temperature $T$ depends on position $(x,y)$, and the position depends on time $t$. Time reaches $T$ by two routes, through $x$ and through $y$, and the rate on the dial is the sum of what each route contributes.
TheoremChain Rule (Case 1): x and y depend on one variable t
Conditions
$z=f(x,y)$ is differentiable
$x=g(t)$ and $y=h(t)$ are differentiable functions of $t$
The rate of z with time is the response of z to x times the speed of x, plus the response of z to y times the speed of y: add the contribution of each route by which t reaches z.
The tree for Case 1. Read $\dfrac{dz}{dt}$ by walking every path from $z$ down to $t$: multiply the labels along each path, then add the paths. Two paths here give the two terms.
Looks like this, but is not
Because $z$ ends up depending on $t$, students often write $\dfrac{\partial z}{\partial t}$ and stop after one term, as if only $x$ carried the time.
There is a single independent variable $t$, so the result is an ordinary derivative $\dfrac{dz}{dt}$, not a partial. And the tree has two branches: dropping the $y$-term is dropping half the answer whenever $y$ actually moves.
How fast is the thermometer climbing?
The temperature near a balloon is $T=x^2+xy$ (with $x$ longitude, $y$ altitude). Right now the balloon is at $(2,1)$, moving with $dx/dt=3$ and $dy/dt=2$. How fast is $T$ changing?
Answer $$\boxed{\dfrac{dT}{dt}=19\ \text{degrees per unit time}}$$
Check
Substitute the path $x=2+3t,\ y=1+2t$ into $T$ and differentiate directly at $t=0$: $2(2)(3)+[3(1)+2(2)]=12+7=19$. Same number, reached without the chain rule.
This is the balloon question from the start, now answered: you never needed a formula for $T$ along the path, only $T$'s partials and the two speeds.
dz/dt for z = x squared y + 3 x y to the fourth at t = 0
Let $z=x^2y+3xy^4$, with $x=\sin 2t$ and $y=\cos t$. Find $dz/dt$ at $t=0$.
With several independent variables, write one partial equation each, summing the tree paths that end at that variable.
Now let the intermediate variables $x,y$ depend on more than one independent variable, say $s$ and $t$. There is nothing new to learn, only more branches to walk: you get one equation per independent variable, each built the same way.
TheoremChain Rule (Case 2): x and y depend on s and t
To differentiate z by one independent variable, walk every path from z down to that variable, multiply the partials along each path, and add the paths; write one such equation for each independent variable.
The tree for Case 2. For $\dfrac{\partial z}{\partial s}$, take the two paths from $z$ that end at $s$; for $\dfrac{\partial z}{\partial t}$, the two that end at $t$. Every equation is 'paths, multiplied along, then summed.'
Looks like this, but is not
Having just met $dz/dt$, it is natural to keep writing $d$'s here too, or to build $\partial z/\partial s$ out of $\partial x/\partial t$.
With two independent variables $s,t$, holding one fixed and moving the other is a partial derivative, so every symbol is a $\partial$. And the wires must match: $\partial z/\partial s$ uses $\partial x/\partial s$ and $\partial y/\partial s$, the rates as $s$ moves, never the $t$-rates. Cross the wires and you have added two unrelated things.
Both partials of z = x squared y with x = s + t, y = s - t
Let $z=x^2y$ with $x=s+t$ and $y=s-t$. Find $\partial z/\partial s$ and $\partial z/\partial t$.
Given
$z=x^2y$
$x=s+t,\ y=s-t$
Find
the two partials $\partial z/\partial s$ and $\partial z/\partial t$
Test at $t=0$: every term has a factor $t$, so $\partial z/\partial s=0$; and indeed at $t=0$, $x=y=0$ for all $s$, making $z=e^0\sin 0=0$ constant in $s$.
Two chain-rule products, then one substitution to return to $s,t$.
Checkpoint
§09.5 — which equation is right●●●○○
The structure is the whole point here, not the arithmetic. Pick the correctly wired equation.
Given
$z=f(x,y)$
$x=g(s,t),\ y=h(s,t)$
Find
(a) Which expression equals $\partial z/\partial t$?
Hint 1/4
Two independent variables are present, so ask first: partial or ordinary derivative, and how many terms.
Hint 2/4
For $\partial z/\partial t$ walk the paths from $z$ that end at $t$; each path multiplies the partials along it, and $t$ is reached through both $x$ and $y$.
Hint 3/4
So the terms use $\partial x/\partial t$ and $\partial y/\partial t$ (the $t$-rates), not the $s$-rates, and there is no direct $z$-to-$t$ link.
Hint 4/4
The answer is $$\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial t}$$.
9.6Slopes without solving: implicit differentiation
Differentiating F = 0 gives dy/dx = -F_x/F_y and, for a surface, partial z by x = -F_x/F_z.
The chain rule pays off at once for equations that tie the variables together without being solved for one of them. Differentiate the whole equation and let the chain rule carry the hidden dependence.
RuleImplicit differentiation formulas
Conditions
$F$ has continuous partials
the relevant denominator partial is non-zero at the point
Write the relation as F equals zero; then the curve's slope is minus the x-partial over the y-partial, and a surface's z-rates are minus the matching partial over the z-partial.
The relation $x^2+y^2=13$ is the level curve $F=0$ of $F(x,y)=x^2+y^2-13$. At $\textcolor{#111}{(2,3)}$ its $\textcolor{#d1690a}{\text{tangent}}$ has slope $-\tfrac{x}{y}=-\tfrac{2}{3}$, exactly what $-F_x/F_y=-\tfrac{2x}{2y}$ gives, with no need to solve for $y$.
Looks like this, but is not
The formula $dy/dx=F_x/F_y$ (no minus sign) looks plausible; after all, we are just taking a ratio of partials.
The minus sign is not optional. Differentiating $F=0$ gives $F_x+F_y\,y'=0$, so $y'$ is $F_x$ moved to the other side and divided out: $-F_x/F_y$. Drop the sign and the tangent tilts the wrong way. In the picture the true slope is negative; $+F_x/F_y=+\tfrac{2}{3}$ would point uphill, straight off the circle.
Slope on the circle x squared plus y squared = 13 at (2, 3)
Find $dy/dx$ for $x^2+y^2=13$ at the point $(2,3)$, without solving for $y$.
Given
$x^2+y^2=13$
point $(2,3)$
Find
the slope $dy/dx$ at the point
SolutionSet F = 0 and take its partials
$$F(x,y)=x^2+y^2-13,\quad F_x=2x,\quad F_y=2y$$
move everything to one side; then two ordinary partials
Solve explicitly to check: $y=\sqrt{13-x^2}$, so $y'=\dfrac{-x}{\sqrt{13-x^2}}=\dfrac{-x}{y}=-\dfrac{2}{3}$ at $(2,3)$. The implicit route skipped the square root.
The slope $-x/y$ is just the direction perpendicular to the radius $(x,y)$, which is why a tangent to a circle is always perpendicular to its radius.
The folium x cubed plus y cubed = 6 x y
Find $dy/dx$ for the folium $x^3+y^3=6xy$, and evaluate it at $(3,3)$.
Given
$x^3+y^3=6xy$
point $(3,3)$
Find
a formula for $dy/dx$, and its value at $(3,3)$
SolutionPartials of F = x cubed + y cubed minus 6 x y
$$F_x=3x^2-6y,\qquad F_y=3y^2-6x$$
differentiate $F=x^3+y^3-6xy$ once with respect to each variable
Symmetry check: the equation is symmetric in $x,y,z$, so $\partial z/\partial y=-\dfrac{y^2+2xz}{z^2+2xy}$ by swapping $x\leftrightarrow y$, which is indeed what the same method gives.
Checkpoint
§09.6 — an implicit partial●●○○○
Set the equation to zero, take two partials, divide with a minus sign.
Given
$x^2+y^2+z^2=14$ defines $z=z(x,y)$
at the point $(1,2,3)$
Find
(a) Find $\partial z/\partial x$ at $(1,2,3)$.
Hint 1/4
You want the surface formula $\partial z/\partial x=-F_x/F_z$; identify $F$ first.
Hint 2/4
Here $F=x^2+y^2+z^2-14$, so $F_x=2x$ and $F_z=2z$.
Hint 3/4
Thus $\partial z/\partial x=-\dfrac{2x}{2z}=-\dfrac{x}{z}$, and at $(1,2,3)$ that is $-\dfrac{1}{3}$.
Direct: $z=(s+t)(s-t)=s^2-t^2$, so $\partial z/\partial s=2s$.
Same function $z=xy$: on the left one clock $t$ drives everything, so the answer is an ordinary $dz/dt$; on the right two dials $s,t$ turn, so the answer is a partial $\partial z/\partial s$ with $t$ held fixed.
How to tell them apart
Count the independent variables. Exactly one gives an ordinary $d$ and a single equation; two or more give a $\partial$ and one equation per independent variable.
Scaffolding comes off
The common skeleton
Identify the dependent variable, the intermediate variables it is built from, and the independent variable(s).
Collect every partial: the derivatives of $z$ in its intermediate variables, and the derivative of each intermediate variable.
For each independent variable, sum over the paths from $z$ down to it, multiplying the labels along each path.
Substitute the given point or expressions, and simplify.
1 · fully worked
dz/dt for z = x squared y with x = t squared, y = t cubed
Find $dz/dt$ at $t=1$ for $z=x^2y$, where $x=t^2$ and $y=t^3$.
Given
$z=x^2y$
$x=t^2,\ y=t^3$
Find
$dz/dt$ at $t=1$
SolutionIdentify and collect
$$z_x=2xy,\quad z_y=x^2$$
the two partials of $z$
$$\frac{dx}{dt}=2t,\quad \frac{dy}{dt}=3t^2$$
one independent variable $t$, so ordinary derivatives
Direct check: $z=(t^2)^2(t^3)=t^7$, so $dz/dt=7t^6=7$ at $t=1$.
2 · you write the reasoning
Same skeleton, an easier function: find $dz/dt$ at $t=2$ for $z=xy$ with $x=t,\ y=t^2$. The steps are written; supply the reason for each before opening it.
$z_x=y$ and $z_y=x$.
reasoning
$z=xy$, so the partial in $x$ leaves $y$, and the partial in $y$ leaves $x$.
$\dfrac{dx}{dt}=1,\ \dfrac{dy}{dt}=2t$.
reasoning
Both intermediate variables depend on the single clock $t$, so these are ordinary derivatives.
$\dfrac{dz}{dt}=y(1)+x(2t)=t^2+2t^2=3t^2$.
reasoning
One independent variable means one equation, summing the two paths $z\to x\to t$ and $z\to y\to t$; then substitute $x=t,\ y=t^2$.
At $t=2$, $\dfrac{dz}{dt}=12$.
reasoning
Finally put $t=2$ into $3t^2$. A direct check agrees: $z=t\cdot t^2=t^3$, $dz/dt=3t^2=12$.
3 · find the buried error
Harder function, and a solution written by someone else. Exactly two of the four steps contain an error. Find both before reading on: the task is to find $dz/dt$ for $z=x^2+y^2$ with $x=e^{t},\ y=e^{-t}$.
Step 1. $z_x=2x$ and $z_y=2y$.
Step 2. $\dfrac{dx}{dt}=e^{t}$ and $\dfrac{dy}{dt}=e^{-t}$.
Direct check: $z=t^2/t=t$, so $dz/dt=1$ for all $t$. The chain rule agrees.
Full exam-style question
Exam level: the polar-coordinate identityexam format
Let $z=f(x,y)$ have continuous partial derivatives, and let $x=r\cos\theta,\ y=r\sin\theta$. Show that $$\left(\dfrac{\partial z}{\partial r}\right)^2+\dfrac{1}{r^2}\left(\dfrac{\partial z}{\partial \theta}\right)^2=\left(\dfrac{\partial z}{\partial x}\right)^2+\left(\dfrac{\partial z}{\partial y}\right)^2.$$
The right side has no $\theta$ in it, so the left side must be rotation-independent too; the cancelling cross terms are exactly what removes the angle. Units also match: both sides are (rate of $z$) squared.
Two chain-rule sums and one trig identity; no properties of $f$ beyond having continuous partials are used.
This is the squared length of the vector of partials written in polar coordinates; you will meet it again as the gradient's length in the next section.
Practice
A · concept 4 questions
1§09.2 — do partials guarantee a plane?●●●○○
One sentence, and it decides whether 'the partials exist' is the same as 'the surface is smooth enough to have a tangent plane'.
Given
$f_x(a,b)$ and $f_y(a,b)$ both exist
Find
(a) True or false: it follows that $f$ is differentiable at $(a,b)$.
Hint 1/4
Try to recall a surface whose partials exist at a point but which has no tangent plane there.
Hint 2/4
Differentiability is stronger than having partials; the theorem needs the partials to be continuous near the point.
Hint 3/4
Take $f=\dfrac{xy}{x^2+y^2}$ with $f(0,0)=0$: both partials are $0$ at the origin, yet $f$ is not even continuous there.
Hint 4/4
So the statement is false: existing partials are not enough.
Check by combining: $\left(\partial z/\partial u\right)^2+\left(\partial z/\partial v\right)^2=4(u^2+v^2)(z_x^2+z_y^2)$, since the cross terms cancel; the factor $4(u^2+v^2)$ is the map's stretch.
3§09.1 — plane then estimate●●●○○
The two halves of 14.4 in one item: build the plane, then read a value off it.
Given
$f(x,y)=\ln(2x+y)$
base point $(-1,3)$
Find
(a) Find the tangent plane at the point above $(-1,3)$.
(b) Use it to estimate $f(-0.9,3.1)$.
Hint 1/4
First check the base value: $2(-1)+3=1$, so $f(-1,3)=\ln 1=0$.
Hint 2/4
The partials are $f_x=\dfrac{2}{2x+y}$ and $f_y=\dfrac{1}{2x+y}$; at $(-1,3)$ they are $2$ and $1$.
Hint 3/4
So the plane is $z=0+2(x+1)+1(y-3)=2x+y-1$; for (b) put $x=-0.9,\ y=3.1$.
Close the page and write, from memory: the tangent-plane formula; what makes a function differentiable and one surface where the partials exist but no tangent plane does; the total differential and why it misses the change by a corner term; the one-parameter chain rule $dz/dt$ and the several-parameter form $\partial z/\partial s$, with the rule for choosing $d$ versus $\partial$; and the two implicit formulas $-F_x/F_y$ and $-F_x/F_z$. Then check against the formula card and mark only the ones you could not produce.
Write the tangent plane to $z=xy^2$ at $(1,2)$ from scratch?
c-tangent-plane
Estimate $\sqrt{(2.02)^2+(1.97)^2}$ by hand, and say why the estimate is trustworthy?
c-linear-approx
Bound the error in a volume $V=xyz$ when each side is off by $0.1$?
c-differentials
Find $dz/dt$ for $z=x^2y$ with $x=\cos t,\ y=\sin t$ without dropping a term?
c-chain-1
Write $\partial z/\partial s$ and $\partial z/\partial t$ for $x=s+t,\ y=s-t$ using a tree?
c-chain-2
Find $\partial z/\partial x$ from $x^2+y^2+z^2=9$, minus sign and denominator correct?
c-implicit
Glossary (8 terms)
tangent planeteğet düzlem
The plane $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$ that touches the surface $z=f(x,y)$ at the point above $(a,b)$.
linearizationdoğrusallaştırma
The tangent plane read as a formula $L(x,y)$ for approximate values of $f$ near the base point.
differentiable (function of two variables)türevlenebilir
A function whose graph has a genuine tangent plane at the point, so the linear approximation error shrinks faster than the distance; guaranteed when $f_x,f_y$ are continuous there.
total differentialtam diferansiyel
The quantity $dz=f_x\,dx+f_y\,dy$, the change in $z$ predicted by the tangent plane over input steps $dx,dy$.
chain rulezincir kuralı
The rule that differentiates a composite by summing, over each intermediate variable, one partial of $z$ times the derivative of that intermediate variable.
tree diagramağaç diyagramı
A picture with the dependent variable on top and the independent ones at the bottom; each path multiplies its partials and the paths to a variable are added.
implicit differentiationkapalı türev
Finding a derivative from a relation $F=0$ without solving for the dependent variable, via $dy/dx=-F_x/F_y$ or $\partial z/\partial x=-F_x/F_z$.
partial derivativekısmi türev
The derivative of a multivariable function with respect to one variable while the others are held fixed.
What comes next
§10 · Directional derivatives, the gradient vector, and maximum and minimum values
The polar identity at the end hinted at it: the two partials $f_x,f_y$ deserve to be bundled into a single vector. Next that bundle gets a name, the gradient, a length, and a direction; it turns the chain rule into a dot product and points the way to the highest and lowest points of a surface.
Sources
James Stewart, Calculus, ninth edition — sections 14.4 and 14.5 The two sections on this week's syllabus line; the tangent plane, differentials, chain rule and implicit formulas follow this book's statements and conventions.
Standard multivariable calculus results The differentiability counterexample and the polar-coordinate identity are classical and appear across standard treatments.