Differentiation rules

Product rule

The product rule generalises to multivariable functions, without modification. So, for a function of two variables, $f(x, y)$ we have:

where, here, the subscript denotes partial differentiation.

Chain rule

We can also use the chain rule but we need to differentiate between two cases, which depend on how many variables we are dealing with. Let us first recall that for the single variable case, for a function $y=f(u)$ where $u=g(x)$, the chain rule gives

We outline the two different cases for a function of two variables below.

Case 1

Consider a function $h(x, y)=f(u(x, y))$ where $f$ is a function of a single variable, i.e. $f=f(u)$ and $u$ is a function of two variables, $u=u(x, y)$. Then this is analogous to the single-variable case shown above but now we have two equations for the two partials of $h$ wrt $x$ and $y$. Applying the chain rule to determine $h_{x}$, we have

and similarly $h_{y}$ is,

Case 2

Consider now a function $h(x, y)=f(u, v)$ where $u=u(x, y)$ and $v=v(x, y)$. Since $h$ is a function of $x$ and $y$, the partial derivatives we want to compute are $h_{x}$ and $h_{y}$, as above. The rate at which $h$ changes with $x$ depends on both $u$ and $v$. The partial derivative of $h$ wrt $x$ in this case therefore is given by the chain rule as follows

and $h_{y}$ is

Note that when differentiating $f$ partially wrt $u(v)$, we treat $v(u)$ as a constant.

The chain rule is used to compute derivatives when changing variables. Suppose we have some quantity $f$ expressed in polar coordinates so that $f(r, \theta)$ is known for any $r$ and $\theta$. Suppose further that we want to differentiate with respect to the Cartesian coordinates, $x$ and $y$ [this often arises in the theory of partial differential equations (PDEs)]. Since $r=r(x, y)$ and $\theta=\theta(x, y)$, we can express $f(r, \theta)$ in terms of $x, y$, as follows

and apply the chain rule.

and for $h_{y}$

In order to obtain $h_{x}$, we need to determine $\partial r / \partial x$ and $\partial \theta / \partial x$ (since the form of $f(r, \theta)$ is known, we can easily obtain $\partial f / \partial r$ and $\partial f / \partial \theta)$. Using $x=r \cos \theta$ and $y=r \sin \theta$, we can express $r$ in terms of $x$ and $y$,

Further, since $\tan \theta=y / x$, taking the inverse yields

where $n$ is an integer.

Laplacian operator

A function that is not 1 - 1 cannot have an inverse unless its domain is restricted. Recall that periodic functions are not $1-1$. For the inverse function of the single-argument $\tan ^{-1} x$ we restrict the domain to $(-\pi / 2, \pi / 2)$; this is an open interval, i.e. the endpoints are not included:

The graph of $y=\tan x$ in $-2 \pi \leq x \leq 2 \pi$. The red part of the graph is in $-\pi / 2<x<\pi / 2$.

The inverse of the red part of graph given by $y=\tan ^{-1} x$; its range is $-\pi / 2<x<\pi / 2$.

Now, let us go back to the two-argument arctan function, $\theta=\tan ^{-1}(y / x)$. We know that $\theta$ is defined in a circle $-\pi<\theta<\pi$. However, $\theta$ is calculated from $\tan ^{-1}(y / x)$ and we know that $-\pi / 2<\tan ^{-1}(y / x)<\pi / 2$. We need to adjust the calculated value of $\theta$ from $\tan ^{-1}(y / x)$ such that it satisfies the sign of the arguments $y$ and $x$ that were used to calculate it. To see this consider the unit circle $(r=1)$ shown below showing the four quadrants of a Cartesian coordinate system.

For $x>0, y>0$, we are in the first quadrant, $y / x>0$ where $0<\theta<\pi / 2$. Indeed, from the graph for $y=\tan ^{-1} x$, for $y / x>0$ we would see that $\theta^{*}$ lies in $(0, \pi / 2)$. It follows that our equation:

yields $\theta=\theta^{*}$, where $\theta^{*}$ represents the angle which satisfies the signs of the arguments $x$ and $y$, and thus we set $n=0$. This is also true when $x>0, y<0$. Moreover, we can show that $x<0, y>0$, we take $n=1$ and for $x<0, y<0$ we take $n=-1$.

However, since we are looking for the partial derivative of $\theta$ wrt $x$ and $y$, the constant $n \pi$ vanishes upon differentiation and does not enter the final result.

From our eqn expressing $r$ in terms of $x$ and $y$,

differentiating partially wrt $x$ gives:

and from Eqn for $\theta$, we have:

where we have used the result,

To see this, let $f(z)=\tan ^{-1} z$ such that

Differentiating implicitly gives,

Using the identity $\sec ^{2} f(z) \equiv \tan ^{2} f(z)+1$,

which is

since $\tan ^{2} f(z)=z^{2}$. Note that as $z \rightarrow \pm \infty, d f / d z \rightarrow 0$.

In polar coordinates, $\partial r / \partial x$ and $\partial \theta / \partial x$ are given by:

Therefore, back in from applying the chain rule to determine $h_{x}$:

which leads to:

The result can be written in the operator form

A similar formula can be obtained for $\partial / \partial y$ using expressions for $\partial r / \partial y$ and $\partial \theta / \partial y$. This is given by ,

To obtain the second partial derivative operators, $\partial^{2} / \partial x^{2}$ and $\partial^{2} / \partial y^{2}$, we partially differentiate wrt $x$ and $y$, respectively. We can then show that their sum simplifies to

This is known as the Laplacian operator and it is extremely useful in the theory of PDEs.

Implicit differentiation

Suppose we have $z=F(x, y)$ with domain $D$ with $y=y(x)$. We put the function in the form:

if the RHS is not zero, we move everything to the left to get it in this form. Now, suppose we want to compute the derivative $\frac{d y}{d x}$. Sometimes we can proceed by solving Eq. (1.30) for $y$ but that is not always possible. Assume we have a point $\left(x_{0}, y_{0}\right) \in D$ such that $F\left(x_{0}, y_{0}\right)=0$ (so that $y\left(x_{0}\right)=y_{0}$ ). Now, along the curve

is a function of $x$ only. We can differentiate with respect to $x$ on both sides:

Using $\frac{d x}{d x}=1$, Eq. (1.32) gives $\frac{d y}{d x}$ as:

provided that $F_{y} \neq 0$. Recall that the subscripts in $F_{x}$ and $F_{y}$ denote partial differentiation with respect to $x$ and $y$, respectively. At the point $\left(x_{0}, y_{0}\right)$, we have:

Equation (1.34) gives the slope of the contour line at the point we started, i.e. in this case $\left(x_{0}, y_{0}\right)$. Of course we can vary the point $\left(x_{0}, y_{0}\right)$ in the domain $D$.

Example 1.11 Let $x y=\ln (2 x+y)$ define a curve on the $x y$-plane. Compute the derivative $\frac{d y}{d x}$ at the point $\left(x_{0}, y_{0}\right)=(0,1)$.

Solution First note that we cannot 'solve for $y$ ' in this case since $y$ appears in the ln function. We proceed by defining:

Note that the domain, $D$ is defined as $2 x+y>0$ since the natural logarithm takes only real positive numbers as the argument. We use Eq. (1.33) to compute $\frac{d y}{d x}$ :

At $(0,1)$, Eq. (1.36) gives $\frac{d y}{d x}=-1$. The same reasoning can be extended to functions of more than two variables, e.g. $\quad w=F(x, y, z)$. Now, $F(x, y, z)=0$ are level surfaces which can be represented as the graph of a function, $z=f(x, y)$. Again, sometimes we can 'solve for $z$ ' and sometimes we cannot. In any case, we can always compute the partial derivatives. So, with $z=f(x, y)$, the level surface is:

Then the derivatives of $F$ are:

Now, from Eqs. (1.38), we can obtain the partial derivatives of the implicit function $f(x, y)$ which defines the level surface of $F$ through $(x, y, z)$. Solving for $f_{x}$ and $f_{y}$ from Eqs. (1.38) yields,