Partial derivatives and the chain rule

This week we carry the concept of the derivative over to multivariable functions. The idea is simple: hold all the other variables constant and take the derivative you know with respect to a single variable. This is called the partial derivative. On top of it we will build two important tools: the chain rule, which computes the derivative when variables depend on other variables, and the implicit differentiation formula, which gives the derivative of functions given by an equation. The exam almost always has a “find all second-order partials”, a chain rule and a related rates question.

Golden rules:

1. When computing $f_x$, $y$ behaves like a number. $y^2$, $\sin y$, $e^y$ are all constant factors.
2. $f_{xy}$ means “first with respect to $x$, then with respect to $y$”. For continuous functions the order does not change the result (Clairaut).
3. In the chain rule, draw a tree: for every path from the dependent variable to the independent variable, multiply the derivatives, and add the paths.
4. In implicit differentiation, do not forget the minus sign at the start of the formula: $\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$.

The partial derivative

Definition.

$f_x(a,b) = \lim_{h\to0}\frac{f(a+h,b)-f(a,b)}{h},\qquad f_y(a,b) = \lim_{h\to0}\frac{f(a,b+h)-f(a,b)}{h}.$

Notations: $f_x = \dfrac{\partial f}{\partial x} = \dfrac{\partial z}{\partial x} = D_xf$.

Geometry. Cut the surface $z = f(x,y)$ with the plane $y = b$. The slope of the resulting curve at the point $(a,b)$ is $f_x(a,b)$. So $f_x$ measures how steeply the surface rises as you walk in the $x$ direction. $f_y$ does the same in the $y$ direction.

Example 1. $f(x,y) = x^3y^2+4xy-y^3$.

$f_x = 3x^2y^2+4y,\qquad f_y = 2x^3y+4x-3y^2.$

At $(1,2)$: $f_x = 12+8 = 20$, $f_y = 4+4-12 = -4$. So at this point the surface rises quickly in the $x$ direction and falls in the $y$ direction.

Example 2 (the chain rule inside). $f(x,y) = \sin(xy^2)$. $f_x = \cos(xy^2)\cdot y^2$. Here $y^2$ is the derivative of the inner function with respect to $x$. $f_y = \cos(xy^2)\cdot2xy$.

Example 3 (quotient). $f(x,y) = \dfrac{x}{x+y}$. $f_x = \dfrac{(x+y)-x}{(x+y)^2} = \dfrac{y}{(x+y)^2}$. $f_y = \dfrac{0-x}{(x+y)^2} = -\dfrac{x}{(x+y)^2}$.

It is the same with three variables: in $f(x,y,z)$, for $f_z$ hold $x$ and $y$ constant.

PITFALL · constant factor or variable? For $f = x^2e^y$, $f_x = 2xe^y$. There is no need to apply the product rule; $e^y$ is constant with respect to $x$. On the other hand $f_y = x^2e^y$, and here $x^2$ is constant. Many people make mistakes in the exam by applying the product rule unnecessarily.

Second-order partial derivatives

You take the partial derivative of the partial derivative again. There are four:

$f_{xx} = \frac{\partial^2f}{\partial x^2},\quad f_{yy} = \frac{\partial^2f}{\partial y^2},\quad f_{xy} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2f}{\partial y\,\partial x},\quad f_{yx} = \frac{\partial^2f}{\partial x\,\partial y}.$

A notation trap: in subscripts the order is left to right ($f_{xy}$: first $x$), while in the $\partial$ notation it is read right to left.

Clairaut’s theorem. If $f_{xy}$ and $f_{yx}$ are continuous around a point, they are equal at that point. This holds for all the functions in the scope of this course, so you should get the same result whichever order you take the mixed derivative in. It is a free way to check your computation.

Example 4. $f(x,y) = xe^{xy}$.

f_x  = e^(xy) + x·y·e^(xy)                 (product rule, x appears twice)
f_xy = x·e^(xy) + [x·e^(xy) + x·y·x·e^(xy)] = 2x e^(xy) + x^2 y e^(xy)
f_y  = x·x·e^(xy) = x^2 e^(xy)
f_yx = 2x e^(xy) + x^2·y·e^(xy)            (the same, Clairaut ✓)
at the point (1,0): f_xy = 2

Laplace’s equation. Functions that satisfy the equation $u_{xx}+u_{yy} = 0$ are called harmonic; heat distribution and electric potential satisfy this equation. $u = e^x\sin y$: $u_{xx} = e^x\sin y$, $u_{yy} = -e^x\sin y$, the sum is $0$.

The chain rule

One parameter. If $z = f(x,y)$, $x = x(t)$, $y = y(t)$, then

$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$

Two parameters. If $x = x(s,t)$, $y = y(s,t)$, then

$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s},\qquad\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}.$

Tree diagram. Put $z$ at the top, the intermediate variables ($x$, $y$) below it, and the independent variables ($s$, $t$) at the bottom. Each path from $z$ to $s$ is a term: multiply the derivatives along the path. Then add all the paths. This rule works for any number of variables.

          z
        /   \
      x       y
     / \     / \
    s   t   s   t
∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s)    ← two paths: z-x-s and z-y-s

Example 5. $z = x^2+y^2$, $x = t^2$, $y = 2t$. $\dfrac{dz}{dt} = 2x\cdot2t+2y\cdot2$. At $t=1$, $x = 1$, $y = 2$: $4+8 = 12$. Check: $z = t^4+4t^2$, $z' = 4t^3+8t = 12$ ✓.

Example 6 (three intermediate variables). $w = xyz$, $x = u+v$, $y = u-v$, $z = uv$. $\dfrac{\partial w}{\partial u} = yz\cdot1+xz\cdot1+xy\cdot v$. At $(u,v) = (2,1)$: $x = 3$, $y = 1$, $z = 2$, the result is $2+6+3 = 11$.

TIP · when do you plug in the values? First write the derivative expression in general. Then compute the values of $x$, $y$ from the values of $s$, $t$. Plug everything in at once at the end. Forgetting to compute the values of the intermediate variables (plugging $s$ in place of $x$) is the most common mistake.

Related rates

The physical face of the chain rule: several quantities change with time, and the rate of change of a composite quantity is asked. Write the formula, apply the chain rule with respect to $t$, plug in the given values. The derivative of a decreasing quantity is entered as negative.

Implicit differentiation

If the equation $F(x,y) = 0$ defines $y$ as a function of $x$, then from the chain rule

$\frac{dy}{dx} = -\frac{F_x}{F_y}\qquad(F_y\neq0).$

With three variables, if $F(x,y,z) = 0$ defines $z$ as a function of $x,y$, then

$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z},\qquad\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}.$

Example 7. $x^3+y^3 = 6xy$ at the point $(3,3)$ (the folium of Descartes). $F = x^3+y^3-6xy$. $F_x = 3x^2-6y = 9$, $F_y = 3y^2-6x = 9$. $\dfrac{dy}{dx} = -1$.

Why the minus? The derivative of the equality $F(x,y(x)) = 0$ with respect to $x$ is $F_x+F_y\dfrac{dy}{dx} = 0$. When you solve, the minus comes out.

Common mistakes