5.14 Chain rule, higher derivatives, and integration

We begin this lecture by studying the differentiability that is the composition of two functions.

Since we are dealing with functions \(f: \mathbb{R}^2 \to \mathbb{R}\), we consider the composition of \(f\) with a function \(\alpha: [a, b] \to \mathbb{R}^2\) whose domain is a closed interval in \(\mathbb{R}\) and take its values in \(\mathbb{R}^2\).

Since \(\alpha: [a, b] \to \mathbb{R}^2\) we write \(\alpha(t) = (x(t), y(t)) \quad \forall t \in [a, b]\)

In particular, one has in each coordinate a function of one variable \(x: [a, b] \to \mathbb{R}\), \(\quad y: [a, b] \to \mathbb{R}\)

Recall that we call such a function a path if it is continuous, that is, for all \(c \in [a, b]\) we have that \(\lim_{t \to c} \alpha(t) = \alpha(c)\) which means that \(\forall \epsilon > 0\), there exists \(\delta > 0\) such that \(||\alpha(t) - \alpha(c)|| < \epsilon\) if \(|t - c| < \delta\)

Exercise: Prove that a function \(\alpha: [a, b] \to \mathbb{R}^2\) with \(\alpha(t) = (x(t), y(t)) \forall t \in [a, b]\) is continuous if and only if the functions \(x: [a, b] \to \mathbb{R}\) and \(x: [a, b] \to \mathbb{R}\) are continuous.

\(\Rightarrow\)) We have \(|t - c| < \delta\) such that \(||\alpha(t) - \alpha(c)|| < \epsilon\), then \(||\left(x(t\right),y\left(t))-\left(x\left(c\right),y\left(c\right)\right)\right .||<\epsilon\)

Then \(\sqrt{\left(x(t\right)-x\left(c\right))^{2}+\left(y\left(t\right)-y\left(c\right)\right)^{2}} <\epsilon\), then we get \(x(t)-x(c)<\varepsilon\) and \(y(t)-y(c)<\varepsilon\)

\(\Leftarrow\)) We have \(x(t)-x(c)<\sqrt{\frac{\varepsilon}{2}}\) and \(y(t)-y(c)<\sqrt{\frac{\varepsilon}{2}}\), then clearly

A path \(\alpha\) is called a differentiable path (\(C^1\)) if it is differentiable for all \(c \in [a, b]\), that is, for all \(c \in [a, b]\) the following limit exists \(\lim_{t \to c} \frac{\alpha(t) - \alpha(c)}{t - c} = \alpha'(c)\)

Write \(\alpha(t) = (x(t), y(t)) \forall t \in [a, b]\). Since \(\alpha(t) - \alpha(c) = (x(t) - x(c), y(t) - y(c))\), it follows that \(\frac{\alpha(t) - \alpha(c)}{t - c} = \left( \frac{x(t) - x(c)}{t - c}, \frac{y(t) - y(c)}{t - c} \right)\)

Hence, the limit exists if and only if both functions \(x: [a, b] \to \mathbb{R}\) and \(y: [a, b] \to \mathbb{R}\) are differentiable and

\(\alpha'(c) = (x'(t), y'(t)) \quad \forall t \in [a, b]\), that is it's differentiable on each coordinate.

Now we study the properties of the function given by the composition of a differentiable path with a differentiable function \(f: \mathbb{R}^2 \to \mathbb{R}\).

So, if we compose such functions

We get a function \(f \circ \alpha: [a, b] \to \mathbb{R} \quad \forall t \in [a, b]\), which is a function of one variable.

Notation: For simplicity, we assume that our interval is \([a, b] = [0, 1]\).

Example

\(f: \mathbb{R}^2 \to \mathbb{R}\), \(f(x, y) = x^2 + y^2\), \(\alpha: [0, 2\pi] \to \mathbb{R}\) where \(\alpha(t) = (\cos(t), \sin(t))\)

\(g(t) = (f \circ \alpha)(t) = f(\alpha(t)) = f(x(t), y(t))\) \(= f(\cos(t), \sin(t))\) \(= \cos^2(t) + \sin^2(t) = 1\)

Theorem

Let \(S \subset \mathbb{R}^2\) be an open set and let \(f: S \to \mathbb{R}\) be a differentiable function.

Let \(x: [0, 1] \to \mathbb{R}\) and \(y: [0, 1] \to \mathbb{R}\) be differentiable functions so that

\[ \alpha(t)=(x(t),y(t))\in S,\forall t\in[0,1] \]

Let \(g(t): [0, 1] \to \mathbb{R}\) be the function given by the composition

\[ g(t)=(f\circ\alpha)(t)=f(x(t),y(t))\quad\forall t\in[0,1] \]

Then \(g\) is differentiable and

\[ g^{\prime}(t_{0})=\frac{\partial f}{\partial x}(x(t_{0}),y(t_{0}))x^{\prime}(t_{0} )+\frac{\partial f}{\partial y}(x(t_{0}),y(t_{0}))y^{\prime}(t_{0}) \]

Proof

Let's \(p=\alpha(t_{0})=(x(t_{0}),y(t_{0}))\) and \(q=\alpha(t_{0}+\delta)=(x(t_0+\delta),y(t_0+\delta))\)

And compute \(x'(t_{0})=\lim_{\delta\to 0}\frac{x(t_0+\delta)-x(t_0)}{\delta},y'(t_{0})=\lim_{\delta\to 0}\frac{y(t_0+\delta)-y(t_0)}{\delta}\)

And \(\lim_{\delta\to 0}\frac{q-p}{\delta}=\lim_{\delta \to 0}\frac{\alpha(t_0+\delta)-\alpha(t_0)}{\delta}=\alpha'(t_0)=(x'(t_0),y'(t_0))\)

By definition, \(g^{\prime}(t_{0})=\lim_{\delta\to0}\frac{g(t_{0}+\delta)-g(t_{0})}{\delta}=\lim_{\delta\to0} \frac{f(\alpha(t_{0}+\delta))-f(\alpha(t_{0}))}{\delta}=\lim_{\delta\to0}\frac{f\left(q\right)-f\left(p\right)}{\delta}\)

Now we observe that we need gradient in the expression. So we add it and minus it

\(=\lim_{\delta\to0}\frac{f\left(q\right)-f\left(p\right)-\nabla f\left(p\right)\left(q-p\right)+\nabla f\left(p\right)\left(q-p\right)}{\delta}=\lim_{\delta\to0}\frac{f\left(q\right)-f\left(p\right)-\nabla f\left(p\right)\left(q-p\right)}{\delta}+\frac{\nabla f\left(p\right)\left(q-p\right)}{\delta}\)

Since \(f\) is differentiable, then \(\lim_{q\to p}\frac{f\left(q\right)-\left(f\left(p\right)+\nabla f\left(p\right)\left(q-p\right)\right)}{\left|\left|q-p\right|\right|} =0=\lim_{\delta\to0}\frac{f\left(q\right)-f\left(p\right)-\nabla f\left(p\right)\left(q-p\right)}{\delta}\)

And \(\lim_{\delta \to 0}\frac{\nabla f(p) \cdot (q - p)}{\delta}\) \(= \lim_{\delta \to 0} \frac{(\frac{\partial f}{\partial x}(p), \frac{\partial f}{\partial y}(p)) \cdot (x(t_0 + \delta) - x(t_0), y(t_0 + \delta) - y(t_0))}{\delta}\)

\(= \lim_{\delta \to 0} \frac{\frac{\partial f}{\partial x}(p)(x(t_0 + \delta) - x(t_0)) + \frac{\partial f}{\partial y}(p)(y(t_0 + \delta) - y(t_0))}{\delta}\) \(= \lim_{\delta \to 0} \left( \frac{\frac{\partial f}{\partial x}(p)(x(t_0 + \delta) - x(t_0))}{\delta} + \frac{\frac{\partial f}{\partial y}(p)(y(t_0 + \delta) - y(t_0))}{\delta} \right)\)

\(= \frac{\partial f}{\partial x}(p) \lim_{\delta \to 0} \frac{x(t_0 + \delta) - x(t_0)}{\delta} + \frac{\partial f}{\partial y}(p) \lim_{\delta \to 0} \frac{y(t_0 + \delta) - y(t_0)}{\delta} = \frac{\partial f}{\partial x}(p)x'(t_0) + \frac{\partial f}{\partial y}(p)y'(t_0)\)

Remark

Another ways of writing the equation above are

\((f \circ \alpha)'(t_0) = \frac{\partial f}{\partial x}(x(t_0), y(t_0))x'(t_0) + \frac{\partial f}{\partial y}(x(t_0), y(t_0))y'(t_0)\)
\((f \circ \alpha)'(t_0) = f_x(x(t_0), y(t_0))x'(t_0) + f_y(x(t_0), y(t_0))y'(t_0)\)
\(\dot{(f \circ \alpha)}(t_{0}) = f_{x}(x(t_{0}), y(t_{0}))\dot{x}(t_{0}) + f_{y}( x(t_{0}),y (t_{0}))\dot{y}(t_{0})\)
\(\displaystyle\frac{d(f \circ \alpha)}{dt}\Big|_{t=t_0}= \frac{\partial f}{\partial x}(x(t_{0}), y (t_{0}))\frac{dx}{dt}\Big|_{t=t_0}+ \frac{\partial f}{\partial y}(x(t_{0} ), y(t_{0})) \frac{dy}{dt}\Big|_{t=t_0}\)
\((f \circ \alpha)'(t_{0}) = \nabla f(x(t_{0}), y(t_{0})) \cdot (x'(t_{0}), y'(t_{0} ))\) \(= \nabla f(x(t_0), y(t_0)) \cdot \alpha'(t_0)\)

Thus, for the equality of functions we write \((f \circ \alpha)' = \nabla f \cdot \alpha'\)

The following corollary is a generalization for more than one variable of the fact that a function with zero derivative has to be constant in an open interval. Here the open interval is replaced by an open and connected set.

Corollary

Let \(S \subset \mathbb{R}^2\) be a domain and let \(f: S \to \mathbb{R}\) be a differentiable function such that

\[ \frac{\partial f}{\partial x}(p)=0=\frac{\partial f}{\partial y}(p)\quad\forall p\in S \]

Then \(f\) is constant in \(S\).

Proof

Since \(S\) is connected and open, then \(S\) is path connected

Actually, we may assume that there exists a diff. path \(\alpha: [0, 1] \to S\) such that \(\alpha(0) = p\), \(\alpha(1) = q\)
We want to prove that \(f\) is constant in \(S\) that is \(f(p) = f(q)\)

We take the composition \(g(t) = f \circ \alpha(t): [0, 1] \to \mathbb{R}\)

\(g\) is diff.
if \(g'(t) = 0 \implies g\) is constant

But \(g'(t) = \frac{\partial f}{\partial x}(t) x'(t) + \frac{\partial f}{\partial y}(t) y'(t)\) \(= 0\) by hypothesis

Then \(g(0)=f(p)=g(1)=f(q)\)

Higher order partial derivatives

Let \(f: S \to \mathbb{R}\) be a differentiable function on an open set \(S \subset \mathbb{R}^2\).

Then the partial derivatives define new functions \(\frac{\partial f}{\partial x}: S \to \mathbb{R}\) and \(\frac{\partial f}{\partial y}: S \to \mathbb{R}\)

One may wonder if these new functions are differentiable or not, and if there is a relation between them.

Let us look at some examples first.

Example

\(f(x,y) = x^2 + y^2\)

Since \(\frac{\partial f}{\partial x}(x,y) = 2x\) and \(\frac{\partial f}{\partial y}(x,y) = 2y\). Then

\(\frac{\partial}{\partial x}(\frac{\partial f}{\partial x}) = 2 = \frac{\partial^2 f}{\partial x^2} = 2\)

\(\frac{\partial}{\partial y}(\frac{\partial f}{\partial x}) = 0 = \frac{\partial^2 f}{\partial y \partial x} = 0\)

\(\frac{\partial}{\partial x}(\frac{\partial f}{\partial y}) = 0 = \frac{\partial^2 f}{\partial x \partial y} = 0\)

\(\frac{\partial}{\partial y}(\frac{\partial f}{\partial y}) = 2 = \frac{\partial^2 f}{\partial y^2} = 2\)
\(f(x,y) = \cos(xy) + x \cos(y)\)

Since \(\frac{\partial f}{\partial x} = -\sin(xy) \cdot y + \cos(y)\) and \(\frac{\partial f}{\partial y} = -\sin(xy) \cdot x - x \sin(y)\). Then

\(\frac{\partial^2 f}{\partial x^2} = -\cos(xy) \cdot y^2\)

\(\frac{\partial^2 f}{\partial y \partial x} = -\cos(xy) \cdot x \cdot y - \sin(xy) - \sin(y)\)

\(\frac{\partial^2 f}{\partial x \partial y} = -\cos(xy) y \cdot x - \sin(xy) - \sin(y)\)

\(\frac{\partial^2 f}{\partial^2 y} = -\cos(xy) \cdot x^2 - x \cos(y)\)

Theorem

Let \(S \subset \mathbb{R}^2\) be an open set and let \(f: S \to \mathbb{R}\) be a differentiable function. Assume that both \(\frac{\partial^2 f}{\partial y \partial x}\) and \(\frac{\partial^2 f}{\partial x \partial y}\) exist and are continuous. Then

\[ \frac{\partial^{2}f}{\partial x\partial y}(p)=\frac{\partial^{2}f}{\partial y\partial x}(p)\quad\forall p\in S \]

Proof

Won't be asked

Integration of functions of two variables.

To introduce the idea of integrable functions of one variable we posed the problem of finding an area below a curve.

Now, if we want to use the same idea for functions in two variables, we observe first that the graph of a function \(f: \mathbb{R}^2 \to \mathbb{R}\) is a surface in \(\mathbb{R}^3\). So, we do not have an area below a curve, but a volume below a surface.

Question: How do we compute such a volume?

Let us now look at an example. Consider the function \(f(x, y) = x^2 + y^2\) and let us try to find the volume below the graph over the closed disk

\(\overline{B((0, 0), 2)}= \{(x, y) \in \mathbb{R}^{2}\mid x^{2}+ y^{2}\leq 4\}\)

Then \(\{y=0\}\cap\overline{B((0,0),2)}=\{x\in\mathbb{R}\mid x^{2}\leq4\}\), then Area\(\displaystyle=\int^{2}_{-2}x^{2}dx\)

If \(y=1,g(x)=x^2+1\), then \(\{y=1\}\cap\overline{B((0,0),2)}=\{x\in\mathbb{R}\mid x^{2}\leq3\}\), then Area\({\displaystyle=\int_{-\sqrt{3}}^{\sqrt{3}}x^2+1dx}\)

If \(y=c\), then \(g(x)=x^2+c^2\). Then \(\{y=c\}\cap\overline{B((0,0),2)}=\{\left(x,c\right)\in\mathbb{R}\mid x^{2}+c^{2} \leq4\}\)

\(=\{x\in \R:-\sqrt{4-c^2}\leq x\leq \sqrt{4-c^2}\}\). Then Area = \(\displaystyle\int^{\sqrt{4-c^{2}}}_{-\sqrt{4-c^{2}}}x^{2}+c^{2}dx\)

To sum infinite things, we take the integral \(\text{Vol}=\displaystyle\int_{2}^{2}\left(\int_{-\sqrt{4-c^{2}}}^{\sqrt{4-c^{2}}}x^2+c^2dx\right) dy\)

If the functions are nice enough, this should represent the volumn