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Multiply along a path and add across paths: the chain rule for one and for several independent variables, the tree diagram that keeps the paths straight, implicit differentiation as a one-line consequence, and the Jacobian matrix whose determinant measures how a change of variables stretches area.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to draw the tree diagram for a composite of several variables, write the chain rule for one and for several independent variables from it, differentiate an implicit relation in one line, build the Jacobian matrix of a change of variables and compute its determinant, and say what a vanishing determinant means for the map.
The one-variable chain rule, $dy/dt = (dy/dx)(dx/dt)$, and partial derivatives from lesson 15. The new ingredient is that a quantity can now reach the variable you are differentiating by more than one route, and the rule has to account for all of them.
An intermediate variable stands between the output and the variable being differentiated; an independent variable is at the bottom of the chain. A tree diagram draws those dependencies, one edge per direct dependence. The Jacobian matrix of a map collects every partial derivative of every output with respect to every input, one row per output; its determinant is the Jacobian determinant, written $\partial(x, y)/\partial(u, v)$.
Case one: one independent variable. If $z = f(x, y)$ with $x = x(t)$ and $y = y(t)$, then
$$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}.$$
There are two routes from $z$ down to $t$ — one through $x$, one through $y$ — and each contributes the product of the derivatives along it. Note which symbols are which: $z$ depends on $t$ alone once the substitution is made, so its derivative is a straight $d$, while the two inner rates are partials.
Case two: several independent variables. If $z = f(x, y)$ with $x = x(s, t)$ and $y = y(s, t)$, then holding $t$ fixed gives
$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s},$$
and the same with $t$ in place of $s$. It is the same rule: every path from the top of the tree to the variable at the bottom, multiplied out and added up.
The tree diagram is the bookkeeping. Write the output at the top, draw an edge down to each variable it depends on directly, and continue until the independent variables are reached. Label each edge with the derivative of the thing above with respect to the thing below. Then: multiply along a path, add across paths. The tree's only job is to stop a path being forgotten, and forgetting one is the commonest error in this lesson.
Implicit differentiation falls out. If $F(x, y) = 0$ defines $y$ as a function of $x$, differentiate both sides with respect to $x$ using the rule: $F_x + F_y\,(dy/dx) = 0$, so
$$\frac{dy}{dx} = -\frac{F_x}{F_y} \qquad (F_y \ne 0).$$
One line replaces the whole implicit-differentiation routine of a first calculus course.
The Jacobian matrix. For a map $(u, v) \mapsto (x, y)$, collect the partials into
$$J = \begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix},$$
one row per output and one column per input. Written this way, the chain rule for composing two maps is exactly matrix multiplication of their Jacobians — the sums of products above are the entries of a matrix product, which is why the rule looks like $(g \circ f)' = g'f'$ once the derivatives are matrices. Its determinant
$$\frac{\partial(x, y)}{\partial(u, v)} = \det J$$
measures how much the map stretches area near a point, and it is the factor that will appear in the change-of-variables formula in lesson 26.
Another way: picture
Draw $z$ at the top of a page with two lines going down to $x$ and $y$, and from each of those a line down to $t$. Two complete routes reach the bottom. Walk each one, multiplying the labels as you go, and add the two walks. A quantity that could be reached three ways would give three terms; the rule never changes, only the count of routes does.
Another way: steps
To differentiate a composite of several variables:
The four partial derivatives of a map $(u, v) \mapsto (x, y)$ could be written as a list. Arranging them as a matrix earns three things a list would not.
Composition becomes multiplication. Apply one map and then another, and the chain rule's sums of products are precisely the entries of the product of the two Jacobians. The one-variable rule multiply the derivatives survives intact — with matrix multiplication, and so with the order of the factors now mattering.
The determinant becomes an area factor. A small square at $(u, v)$ with sides $du$ and $dv$ is carried to a small parallelogram whose edges are the columns of $J$ scaled by $du$ and $dv$. From unit 1, the area of a parallelogram spanned by two vectors is the size of a cross product — and in two dimensions that is exactly $|\det J|$. So $|\det J|\,du\,dv$ is the image's area, which is the entire content of the change-of-variables formula.
The sign becomes an orientation. A negative determinant means the map reflects the plane, swapping clockwise for anticlockwise. That is why the change-of-variables formula carries an absolute value, and why the orientation conventions of unit 5 have to be stated rather than assumed.
Missing a path. If $z = f(x, y)$ and both $x$ and $y$ depend on $t$, then $dz/dt$ has two terms. Writing only $(\partial z/\partial x)(dx/dt)$ is the single commonest error here, and drawing the tree prevents it.
Substituting too early. Putting $t = 2$ into $x(t)$ before differentiating leaves a constant, whose derivative is zero. Differentiate, then substitute.
Confusing $\partial z/\partial x$ with $dz/dx$. In case two, $z$ depends on $s$ and $t$ through $x$ and $y$; $\partial z/\partial x$ holds $y$ fixed, while $\partial z/\partial s$ holds $t$ fixed. They are different questions, and the notation only distinguishes them if the variables being held are named when there is any doubt.
Transposing the Jacobian. Rows are outputs and columns are inputs. The transpose is a perfectly good matrix and the wrong one, and for a non-square map it will not even multiply. A third coordinate is not a third term bolted onto a two-dimensional formula. The cross product exists only in space, a plane is fixed by a direction rather than by a slope, and a curve in space can bend without ever leaving a surface. Checking a new formula against the plane is worth doing; assuming the plane's formula still governs is how most of the errors in this course begin.
The one-variable chain rule had a single chain, so it is natural to write $dz/dt = (\partial z/\partial x)(dx/dt)$ and stop. That is not a slightly incomplete answer; it is the derivative of a different problem — the one in which $y$ happens to be constant.
The way to keep it straight is to insist on the tree before any differentiating starts. Draw the output, draw an edge to everything it depends on directly, and keep going down. Then count the complete routes to the variable at the bottom. Two routes means two terms; three means three. The count is settled by the diagram rather than by how the formula looks, and that is the point of drawing it.
The same discipline settles the notation. $\partial z/\partial x$ is a label on one edge — the rate of $z$ against $x$ with the other middle variable held. $dz/dt$ is a property of the whole tree. They are quantities of different kinds, and writing $d$ where a $\partial$ belongs usually means a path has gone missing.
Let $z = x^2y$ with $x = \cos t$ and $y = \sin t$. The edges carry $\partial z/\partial x = 2xy$, $\partial z/\partial y = x^2$, $dx/dt = -\sin t$ and $dy/dt = \cos t$.
Four edges, two complete routes.
So $dz/dt = 2xy(-\sin t) + x^2(\cos t) = -2\cos t \sin^2 t + \cos^3 t$.
Multiply along each path, then add.
Substituting first gives $z = \cos^2 t \sin t$, and the product rule gives the same thing. Both routes work here because the substitution was easy; the chain rule earns its keep when $f$ is given by a table or by a name rather than by a formula, and there is nothing to substitute into.
The rule and the substitution agree, as they must.
The map $(r, \theta) \mapsto (x, y) = (r\cos\theta, r\sin\theta)$ has $\partial x/\partial r = \cos\theta$, $\partial x/\partial\theta = -r\sin\theta$, $\partial y/\partial r = \sin\theta$ and $\partial y/\partial\theta = r\cos\theta$.
One row per output variable.
Its determinant is $\cos\theta \cdot r\cos\theta - (-r\sin\theta)\cdot\sin\theta = r(\cos^2\theta + \sin^2\theta) = r$.
Cross-multiply the diagonals and subtract.
So a small patch $dr\,d\theta$ becomes a patch of area $r\,dr\,d\theta$ — which is the factor lesson 23 will insist on, derived here rather than asserted. It vanishes at $r = 0$, where the whole circle of angles is crushed onto a single point and the map stops being reversible.
The famous extra $r$ is a Jacobian determinant.
Let $z = x^2 + y^2$ with $x = s + t$ and $y = s - t$. The tree has $z$ on top, $x$ and $y$ below, and both $s$ and $t$ under each of those.
Four edges from the middle row down.
For $\partial z/\partial s$: the two paths give $2x \cdot 1$ and $2y \cdot 1$, so $\partial z/\partial s = 2x + 2y = 2(s + t) + 2(s - t) = 4s$.
For $\partial z/\partial t$: the paths give $2x \cdot 1$ and $2y \cdot (-1)$, so $\partial z/\partial t = 2x - 2y = 4t$. Substituting first gives $z = 2s^2 + 2t^2$, which agrees — and shows what the change of variables achieved: a function that mixed $x$ and $y$ has separated into one term per new variable.
A change of variables sends $(u, v)$ to $x = 3u + 5v^2$ and $y = 3u^2 + 4v$. Write its Jacobian matrix at the point $u = 3$, $v = 2$.
This task has no paper form; do it on a device.
For $z = f(x, y)$ with $x = x(t)$ and $y = y(t)$, put the chain rule calculation in order. Suppose $\partial z/\partial x = 6$ at the point of interest.
Number the steps in order (write the number in the box):
Let $z = 6x + 5y$ with $x = t^2$ and $y = t^3$. What is $dz/dt$ at $t = 2$?
Answer:
A change of variables sends $(u, v)$ to $x = 4u + 3v$ and $y = u + 3v$. What is the determinant of its Jacobian?
Answer:
A change of variables sends $(u, v)$ to $x = 5u + 2v^2$ and $y = 2u^2 + 5v$. Write its Jacobian matrix at the point $u = 2$, $v = 1$.
This task has no paper form; do it on a device.
A change of variables has Jacobian determinant $2uv$, which is zero all along the line $u = 0$. What does that say about the map near a point of that line?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $z = f(x, y)$ with $x = x(t)$ and $y = y(t)$, put the chain rule calculation in order. Suppose $\partial z/\partial x = 9$ at the point of interest.
Number the steps in order (write the number in the box):
You can find every path from an output down to the variable you are differentiating, multiply along each and add across them, and read a Jacobian determinant as the factor by which a map stretches area. Next: the direction in which a function climbs fastest, and the vector that names it.
10. Your turn: two independent variables, step 3