📖 Applications of derivatives

📖 Applications of derivatives#

⏱ | words

Univariate optimization#

Let \(f \colon [a, b] \to \mathbb{R}\) be a differentiable (smooth) function

\([a, b]\) is all \(x\) with \(a \leq x \leq b\)
\(\mathbb{R}\) is “all numbers”
\(f\) takes \(x \in [a, b]\) and returns number \(f(x)\)
derivative \(f'(x)\) exists for all \(x\) with \(a < x < b\)

Definition

A point \(x^* \in [a, b]\) is called a

maximizer of \(f\) on \([a, b]\) if \(f(x^*) \geq f(x)\) for all \(x \in [a,b]\)
minimizer of \(f\) on \([a, b]\) if \(f(x^*) \leq f(x)\) for all \(x \in [a,b]\)

Example

Let

\(f(x) = -(x-4)^2 + 10\)
\(a = 2\) and \(b=8\)

Then

\(x^* = 4\) is a maximizer of \(f\) on \([2, 8]\)
\(x^{**} = 8\) is a minimizer of \(f\) on \([2, 8]\)

_images/c7a242957c4475c13de9f658c6849ab0731b298d76a57f60d401b9af965c4f3d.png — Fig. 31 Maximizer on \([a, b] = [2, 8]\) is \(x^* = 4\)#

_images/0edc41437ce2ba52a68629337e245893bfbaed65137f325bda6f43e7b3062389.png — Fig. 32 Minimizer on \([a, b] = [2, 8]\) is \(x^{**} = 8\)#

The set of maximizers/minimizers can be

empty
a singleton (contains one element)
finite (contains a number of elements)
infinite (contains infinitely many elements)

Example: infinite maximizers

\(f \colon [0, 1] \to \mathbb{R}\) defined by \(f(x) =1\)
has infinitely many maximizers and minimizers on \([0, 1]\)

Example: no maximizers

The following function has no maximizers on \([0, 2]\)

\[\begin{split} f(x) = \begin{cases} x^2 & \text{ if } x < 1 \\ 1/2 & \text{ otherwise} \end{cases} \end{split}\]

_images/b4991b4f20b0d9ea4f5535aeb65ae28b0a312cd281d789194f851d8dfcd7ac57.png — Fig. 33 No maximizer on \([0, 2]\)#

Definition

Point \(x\) is called interior to \([a, b]\) if \(a < x < b\)

The set of all interior points is written \((a, b)\)

We refer to \(x^* \in [a, b]\) as

interior maximizer if both a maximizer and interior
interior minimizer if both a minimizer and interior

Finding optima#

Definition

A stationary point of \(f\) on \([a, b]\) is an interior point \(x\) with \(f'(x) = 0\)

_images/stationary1.png — Fig. 34 Interior maximizers/minimizers are stationary points.#

_images/stationary2.png — Fig. 35 Not all stationary points are maximizers!#

Fact

If \(f\) is differentiable and \(x^*\) is either an interior minimizer or an interior maximizer of \(f\) on \([a, b]\), then \(x^*\) is stationary

Sketch of proof, for maximizers:

\[ f'(x^*) = \, \lim_{h \to 0} \, \frac{f(x^* + h) - f(x^*)}{h} \qquad \text{(by def.)} \]

\[ \Rightarrow f(x^* + h) \approx f(x^*) + f'(x^*) h \qquad \text{for small } h \]

If \(f'(x^*) \ne 0\) then exists small \(h\) such that \(f(x^* + h) > f(x^*)\)

Hence interior maximizers must be stationary — otherwise we can do better

Fact

Previous fact \(\implies\)

\(\implies\) any interior maximizer stationary
\(\implies\) set of interior maximizers \(\subset\) set of stationary points
\(\implies\) maximizers \(\subset\) stationary points \(\cup \{a\} \cup \{b\}\)

Algorithm for univariate problems

Locate stationary points
Evaluate \(y = f(x)\) for each stationary \(x\) and for \(a\), \(b\)
Pick point giving largest \(y\) value

Minimization: same idea

Example

Let’s solve

\[ \max_{-2 \leq x \leq 5} f(x) \quad \text{where} \quad f(x) = x^3 - 6x^2 + 4x + 8 \]

Steps

Differentiate to get \(f'(x) = 3x^2 - 12x + 4\)
Solve \(3x^2 - 12x + 4 = 0\) to get stationary \(x\)
Discard any stationary points outside \([-2, 5]\)
Eval \(f\) at remaining points plus end points \(-2\) and \(5\)
Pick point giving largest value

from sympy import *
x = Symbol('x')
points = [-2, 5]
f = x**3 - 6*x**2 + 4*x + 8
fp = diff(f, x)
spoints = solve(fp, x)
points.extend(spoints)
v = [f.subs(x, c).evalf() for c in points]
maximizer = points[v.index(max(v))]
print("Maximizer =", str(maximizer),'=',maximizer.evalf())

---------------------------------------------------------------------------
ModuleNotFoundError                       Traceback (most recent call last)
Cell In[2], line 1
----> 1 from sympy import *
      2 x = Symbol('x')
      3 points = [-2, 5]

ModuleNotFoundError: No module named 'sympy'

Shape Conditions and Sufficiency#

When is \(f'(x^*) = 0\) sufficient for \(x^*\) to be a maximizer?

One answer: When \(f\) is concave

(Full definition deferred)

Sufficient conditions for concavity in one dimension

Let \(f \colon [a, b] \to \mathbb{R}\)

If \(f''(x) \leq 0\) for all \(x \in (a, b)\) then \(f\) is concave on \((a, b)\)
If \(f''(x) < 0\) for all \(x \in (a, b)\) then \(f\) is strictly concave on \((a, b)\)

Example

\(f(x) = a + b x\) is concave on \(\mathbb{R}\) but not strictly
\(f(x) = \log(x)\) is strictly concave on \((0, \infty)\)

When is \(f'(x^*) = 0\) sufficient for \(x^*\) to be a minimizer?

One answer: When \(f\) is convex

(Full definition deferred)

Sufficient conditions for convexity in one dimension

Let \(f \colon [a, b] \to \mathbb{R}\)

If \(f''(x) \geq 0\) for all \(x \in (a, b)\) then \(f\) is convex on \((a, b)\)
If \(f''(x) > 0\) for all \(x \in (a, b)\) then \(f\) is strictly convex on \((a, b)\)

Example

\(f(x) = a + b x\) is convex on \(\mathbb{R}\) but not strictly
\(f(x) = x^2\) is strictly convex on \(\mathbb{R}\)

Sufficiency and uniqueness with shape conditions#

Fact

For maximizers:

If \(f \colon [a,b] \to \mathbb{R}\) is concave and \(x^* \in (a, b)\) is stationary then \(x^*\) is a maximizer
If, in addition, \(f\) is strictly concave, then \(x^*\) is the unique maximizer

Fact

For minimizers:

If \(f \colon [a,b] \to \mathbb{R}\) is convex and \(x^* \in (a, b)\) is stationary then \(x^*\) is a minimizer
If, in addition, \(f\) is strictly convex, then \(x^*\) is the unique minimizer

Example

A price taking firm faces output price \(p > 0\), input price \(w >0\)

Maximize profits with respect to input \(\ell\)

\[ \max_{\ell \ge 0} \pi(\ell) = p f(\ell) - w \ell, \]

where the production technology is given by

\[ f(\ell) = \ell^{\alpha}, 0 < \alpha < 1. \]

Evidently

\[ \pi'(\ell) = \alpha p \ell^{\alpha - 1} - w, \]

so unique stationary point is

\[ \ell^* = (\alpha p/w)^{1/(1 - \alpha)} \]

Moreover,

\[ \pi''(\ell) = \alpha (\alpha - 1) p \ell^{\alpha - 2} < 0 \]

for all \(\ell \ge 0\) so \(\ell^*\) is unique maximizer.

Implicit differentiation#

WARNING

This section of the lecture notes is still under construction. It will be ready before the lecture.

L’Hopital’s rule#

Consider the function \(f(x)=\frac{g(x)}{h(x)}\).

Definition

If \(\lim _{x \rightarrow x_{0}} g(x)=\lim _{x \rightarrow x_{0}} h(x)=0\), we call this situation an “\(\frac{0}{0}\)” indeterminate form

If \(\lim _{x \rightarrow x_{0}} g(x)= \pm \infty\), \(\lim _{x \rightarrow x_{0}} h(x)= \pm \infty\), we call this situation an “\(\frac{ \pm \infty}{ \pm \infty}\)” indeterminate form

Fact: L’Hopital’s rule for limits of indeterminate form

Suppose that \(g(x)\) and \(h(x)\) are both differentiable at all points on a non-empty interval \((a, b)\) that contains the point \(x=x_{0}\), with the possible exception that they might not be differentiable at the point \(x_{0}\).

Suppose also that \(h^{\prime}(x) \neq 0\) for all \(x \in(a, b)\).

Further, assume that for \(f(x)=\frac{g(x)}{h(x)}\) we have either “\(\frac{0}{0}\)” or “\(\frac{ \pm \infty}{ \pm \infty}\)” indeterminate form.

Then if \(\lim _{x \rightarrow x_{0}}\frac{g^{\prime}(x)}{h^{\prime}(x)}\) exists, then

\[ \lim _{x \rightarrow x_{0}} f(x) = \lim _{x \rightarrow x_{0}}\frac{g(x)}{h(x)} = \lim _{x \rightarrow x_{0}}\frac{g^{\prime}(x)}{h^{\prime}(x)} \]

all four conditions have to hold for the L’Hopital’s rule to hold!

Example

Note that in the examples below we have “\(\frac{0}{0}\)” indeterminacy, and both enumerator and denominator functions are differentiable around \(x=0\)

\[ \lim_{x \to 0} \frac{\ln(x+1)}{x} = \lim_{x \to 0} \frac{\frac{1}{x+1}}{1} = \lim_{x \to 0} \frac{1}{x+1} = 1 \]

\[ \lim_{x \to 0} \frac{e^x-1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1 \]

Example

L’Hopital’s rule is not applicable for \(\lim_{x \to 0} \frac{x}{\ln(x)}\) because neither “\(\frac{0}{0}\)” nor “\(\frac{ \pm \infty}{ \pm \infty}\)” indeterminacy form applies.

Elasticity#

Elasticity and in particular point elasticity is a concept that is widely used in economics for measuring responsiveness of economic quantities (demand, labor supply, etc.) to changes in other economic quantities (prices, wages, etc.)

Definition (in words)

Suppose that \(y=f(x)\). The elasticity of \(y\) with respect to \(x\) is defined to be the percentage change in \(y\) that is induced by a small (infinitesimal) percentage change in \(x\).

So, what is a percentage change in \(x\)?

\(x\) to \(x+h\), making \(h\) the absolute change in \(x\)
\(x\) to \(x + rx\), making \((x+rx)/x = 1+r\) the relative change in \(x\), and \(r\) is the relative change rate
percentage change in \(x\) is the relative change rate expressed in percentage points

We are interested in infinitesimal percentage change, therefore in the limit as \(r \to 0\) (irrespective of the units of \(r\))

What is the percentage change in \(y\)?

Answer: rate of change of \(f(x)\) that can be expressed as

\[ \frac{f(x+rx)}{f(x)} -1 = \frac{f\big(x(1+r)\big)-f(x)}{f(x)} \]

Let’s consider the limit

\[ \lim_{r \to 0} \frac{\frac{f\big(x(1+r)\big)-f(x)}{f(x)}}{r} = \frac{1}{f(x)} \lim_{r \to 0} \frac{f\big(x(1+r)\big)-f(x)}{r} = \]

Define a new function \(g\) such that \(f(x) = g\big(\ln(x)\big)\) and \(f\big(x(1+r)\big) = g\big(\ln(x) + \ln(1+r)\big)\)

Continuing the above, and noting that \(\ln(1+r) \to 0\) as \(r \to 0\) we have

\[\begin{split} \begin{array}{l} = \frac{1}{f(x)} \lim_{r \to 0} \frac{g\big(\ln(x) + \ln(1+r)\big)-g\big(\ln(x)\big)}{r} = \\ = \frac{1}{f(x)} \lim_{r \to 0} \frac{g\big(\ln(x) + \ln(1+r)\big)-g\big(\ln(x)\big)}{\ln(1+r)} \frac{\ln(1+r)}{r} = \\ = \frac{1}{f(x)} \left[ \lim_{r \to 0} \frac{g\big(\ln(x) + \ln(1+r)\big)-g\big(\ln(x)\big)}{\ln(1+r)} \lim_{r \to 0} \frac{\ln(1+r)}{r} \right] =\\= \frac{1}{f(x)} \frac{d g(\ln(x))}{d \ln(x)} = \frac{1}{f(x)} \frac{d f(x)}{d \ln(x)} = \frac{x}{f(x)} \frac{d f(x)}{d x} \end{array} \end{split}\]

See the example above showing that \(\lim_{r \to 0} \frac{\ln(1+r)}{r} = 1\).

In the last step, to make sense of the expression \(\frac{d f(x)}{d \ln(x)}\), imagine that \(x\) is a function of some other variable we call \(\ln(x)\). Then we can use the chain rule and the inverse function rule to show that

\[ \frac{d f(x)}{d \ln(x)} = \frac{d f(x)}{d x} \frac{dx}{d \ln(x)} = \frac{d f(x)}{d x} \frac{1}{\frac{1}{x}} = x \frac{d f(x)}{d x} \]

Note, in addition, that by the chain rule the following also holds

\[ \frac{d \ln f(x)}{d \ln(x)} = \frac{1}{f(x)} \frac{d f(x)}{d \ln(x)} \]

Putting everything together, we arrive at the following equivalent expressions for elasticity.

Definition (exact)

The (point) elasticity of \(f(x)\) with respect to \(x\) is defined to be the percentage change in \(y\) that is induced by a small (infinitesimal) percentage change in \(x\), and is given by

\[ \epsilon^{f}_{x}(x)= \lim_{r \to 0} \frac{\frac{f\big(x(1+r)\big)}{f(x)}-1}{r} = \frac{\frac{d f(x)}{d x}}{\frac{f(x)}{x}} = \frac{x}{f(x)} \frac{d f(x)}{d x} = \frac{d \ln f(x)}{d \ln x} \]

Example

Suppose that we have a linear demand curve of the form

\[ Q = a − bP \]

where \(a > 0\) and \(b > 0\) The slope of this demand curve is equal to \((−b)\) As such, the point own-price elasticity of demand in this case is given by

\[\begin{split} \begin{align*} \epsilon^D_P (P) &= \left( \frac{P}{a - bP} \right) (-b) \\ &= \left( \frac{-bP}{a - bP} \right) \end{align*} \end{split}\]

Note that

\[\begin{split} \begin{align*} | \epsilon^D_P (P) | &= 1 \\ \iff | \frac{-bP}{a - bP} | &= 1 \\ \iff \frac{bP}{a - bP} &= 1 \\ \iff bP &= a - bP \\ \iff 2bP &= a \\ \iff P &= \frac{a}{2b} \\ \end{align*} \end{split}\]

More generally, we can show that, for the linear demand curve above, we have

\[\begin{split} | \epsilon^D_P (P) | = \begin{cases} \in (0, 1) \;\; \text{ if } P < \frac{a}{2b} \\ = 1 \;\; \text{ if } P = \frac{a}{2b} \\ \in (1, \infty) \;\; \text{ if } P > \frac{a}{2b} \\ \end{cases} \end{split}\]

Since the demand curve above is downward sloping, this means that

\[\begin{split} \epsilon^D_P (P) = \begin{cases} \in (-1, 0) \;\; \text{ if } P < \frac{a}{2b} \\ = -1 \;\; \text{ if } P = \frac{a}{2b} \\ \in (-\infty, -1) \;\; \text{ if } P > \frac{a}{2b} \\ \end{cases} \end{split}\]

Example

Suppose that we have a demand curve of the form

\[ Q = a/P \]

where \(a > 0\). Computing the point elasticity of demand at price \(P\) we have

\[\begin{split} \begin{align*} \epsilon^D_P (P) &= \left( \frac{P}{a/P} \right) \left( \frac{-a}{P^2} \right) \\ &= \frac{P^2}{a}\frac{-a}{P^2} = -1 \end{align*} \end{split}\]

Constant own-price elasticity!

Alternatively we could compute this elasticity as

\[\begin{split} Q &= a/P = a P^{-1} \quad \Leftrightarrow \quad \ln Q = \ln a - \ln P \\ \epsilon^D_P (P) &= \frac{d \ln Q}{d \ln P} = -1 \end{split}\]

Newton-Raphson (Newton) method#

Classic method for fast numerical solving the equation of the form \(f(x)=0\)

Equation solving
Finding maximum/minimum based on FOC, then \( f(x)=Q'(x) \)

Algorithm:

Start with some good guess \( x_0 \) not too far from the solution
Newton step: \( x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)} \)
Iterate until convergence in some metric

Derivation based on the Taylor series#

\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(x_0)}{k!} (x-x_0)^k \]

Take first two terms, assume \( f(x) \) is solution, and let \( x_0=x_i \) and \( x=x_{i+1} \)

\[ 0 = f(x) = f(x_i) + f'(x_i) (x_{i+1}-x_i) \quad \Rightarrow \quad x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)} \]

Example

See the working example in Jupyter Notebook

📖 Applications of derivatives

Contents

📖 Applications of derivatives#

Univariate optimization#

Finding optima#

Shape Conditions and Sufficiency#

Sufficiency and uniqueness with shape conditions#

Implicit differentiation#

L’Hopital’s rule#

Elasticity#

Newton-Raphson (Newton) method#

Derivation based on the Taylor series#