Slopes of curves
- We will first focus on the geometrical relevance off the notion of derivative: the slope of a curve
- Important: Under most circumstances, we always refer to the slope of a curve at a given point of the curve
Example
- Take a function \(y=f\left(x\right)\). Draw it. Draw a straight line tangent to f(x) at point \(\left(a,\ f\left(a\right)\right)\).
- We refer to the slope of the tangent as the derivative of f(x) at point a.
- The notation for this derivative is: \(f^{'}(a)\) (f prime of a)
Tangents
- Draw the graph from p. 165
- The formula for the slope of this curve is given by: \(m_{PQ}=\frac{f(a+h)-f(a)}{h}\), where the numerator is the vertical distance between the two points and the denominator is the horizontal distance between the two
- This is also referred to as Newton quotient
- Important: what happens when h=0?
- Nevertheless, we want to find a formula for the tangent when h tends towards zero:
- We thus have the following definition of a derivative: \(f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\)
- We can now use the point-slope formula to derive the equation for our tangent: \(y-f(a)=f'(a)(x-a)\)
Example
Derive the derivative of \(f(x)=x^2\) using the method above
\(f(x)=x^2\) implies \(f(a+h)=(a+h)^2\) and \(f(a)=a^2\).
Substituting in the formula for the derivative yields
\(f'(a)=\lim_{h\to 0}\frac{(a+h)^2-a^2}{h}\)
Basic arithmetic yields
\(\)\(f'(a)=\lim_{h\to 0}\frac{a^2+2ah+h^2-a^2}{h}=\lim_{h\to 0}\frac{2ah+h^2}{h}=\lim_{h\to 0}\frac{2a+h}{1}=2a\)
- If f is a relatively simple function, we can find the derivative by following the following rules:
- Add \(h\neq 0\) to \(a\) and compute \(f(a+h)\)
- Compute \(f(a+h)-f(a)\)
- Form the Newton quotient
- Simplify the resulting formula as much as possible, making sure not to have \(h\) in the denominator
- Take the limit as h tends to 0
Example
Use these rules to find the derivative of \(f(x)=x^3\)
Some notational issues
- There are a wide array of notations when it comes to derivatives (this is because Newton and Leibniz "discovered" calculus): \(\frac{dy}{dx};\ \frac{df\left(x\right)}{dx};\ \frac{d}{dx}f\left(x\right)\)
Increasing and decreasing functions
- If \(f\left(x_2\right)\ge f\left(x_1\right)\) whenever \(x_2>x_1\), the function is increasing
- If \(f\left(x_2\right)\le f\left(x_1\right)\) whenever \(x_2>x_1\), the function is decreasing
- If \(f(x_2)>f(x_1)\) whenever \(x_2>x_1\), the function is strictly increasing
- If \(f(x_2)<f(x_1)\) whenever \(x_2>x_1\), the function is strictly decreasing
We can reformulate this conditions in terms of the function's derivative:
- If \(f'(x) \ge 0\) for all \(x\) in the interval I, the function is increasing over I (and vice-versa)
- If \(f'(x)\le 0\) for all \(x\) in the interval I, the function is decreasing over I (and vice-versa)
- \(\)If \(f'(x)=0\) for all \(x\) in the interval I, the function is constant over I (and vice-versa)
Examples
- Find whether the function \(f(x)=\frac{1}{2}x^2-2\) is increasing or decreasing
- Find whether the function \(f(x)=-\frac{1}{3}x^3+2x^2-3x+1\) is increasing or decreasing
Rates of change
- In economics, we focus on a precise interpretation of the notion of derivative: the derivative is seen as the rate of change of a variable
Rate of change interpretation
- Take a function \(f(x)\). Newton's quotient can then be interpreted as the average rate of change in \(f(x)\). \(f(x)\) changes at every point over the interval. This change is expressed by the difference in the numerator. Dividing this change by the change in \(x\) (e.g., \(x\)) is the average rate of change
- Thus, using the interpretation of the Newton's formula from above, the derivative can be interpreted as \(\)the instantaneous rate of change of the function
- We can now derive the relative rate of change or proportional rate of change: \(\frac{f'(a)}{f(a)}\)
Example
Consider a firm producing some commodity in a given period, with
- \(C(x)\) being the cost function
- \(R(x)\) being the revenue function
- \(\pi(x)=R(x)-C(x)\) is the profit function
We can use Newton's quotient to find the marginal cost of the firm:
\(C'(x)=\lim_{h\to 0}\frac{C(x+h)-C(h)}{x}\)
For an \(h\) small enough, we can write
\(C'(x)\simeq \frac{C(x+h)-C(x)}{h}\)
For \(h=1\) we obtain
\(C'(x)\simeq C(x+1)-C(x)\)
Thus, the interpretation of marginal cost is (approximately) the increase in cost due to an increase in output by one unit
Differentiability and empirical functions
Economists use derivatives to study the rate of change of a wide variety of variables. In some case, this is mathematically improper, as the variable of interest is discrete and not continuous and we cannot define a derivative for discrete functions
A dash of limits
- We say that \(f(x)\) has the number A as its limit as \(x\) tends to \(a\), if \(x\) tends to A as \(x\) tends to (but is not equal to) \(a\):
\(\lim_{x\to a} f(x)=A\) or \(f(x) \to A\) as \(x\to a\)
- We say that the limit does not exist if \(f(x)\) does not tend to any specific value as \(x\) tends to \(a\)
- Important: writing \(\lim_{x\to a}f(x)=A\) means that we can make \(f(x)\) as close to \(A\) as we want for all \(x\) sufficiently close to (but not equal to) \(a\)
Rules of limits
If \(\lim_{x\to a}f(x)=A\) and \(\lim_{x\to a}g(x)=B\), then
- \(\lim_{x\to a}(f(x)\pm g(x))=A\pm B\)
- \(\lim_{x\to a}(f(x)\cdot g(x))=A\cdot B\)
- \(\lim_{x\to a}\frac{f(x)}{g(x)}=A/B, B\neq 0\)
- \(\lim_{x\to a}(f(x))^r=A^r\)
Examples
- \(\lim_{x\to 2}(x^2+5x)\)
- \(\lim_{x\to 4}\frac{2x^{3/2}-\sqrt{x}}{x^2-15}\)
- \(\lim_{x\to a}Ax^n\)
Simple rules of differentiation
- If the limit \(f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\) exists, then we say that the function f is differentiable at
Some rules
- If the function is a constant, its derivative is always equal to zero, or \(f(x)=A \Longrightarrow f'(x)=0\)
- What is the intuition behind this?
- We can extend this to derive the following rule: the derivative of a function containing an additive constant is given by
\(y=A+f(x) \Longrightarrow y'=f'(x)\)
- If the constant multiplies the function, then the formula for the derivative is
\(y=Af(x) \Longrightarrow y'=Af'(x)\)
Proof
\(f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\)
Now write
\(g(a)=Af(a)\)
Which implies
\(g'(a)=\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}\)
Substituting from above yields
\(g'(a)=\lim_{h\to 0}\frac{Af(a+h)-Af(a)}{h} \Longrightarrow g'(a)=\lim_{h\to 0}A\frac{f(a+h)-f(a)}{h}\)
Or
\(g'(a)=A\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=Af'(a)\)
QED
Examples
Find the derivative of the following functions
- \(y=5+f(x)\)
- \(y=f(x)-\frac{1}{2}\)
- \(y=4f(x)\)
- \(y=\frac{Af(x)+B}{C}\)
The power rule
\(f(x)=x^a\Longrightarrow f'(x)=ax^{a-1}\), wehere \(a\) is constant
Examples
Find the derivative of the following functions
- \(y=x^5\)
- \(y=3x^8\)
- \(y=\frac{x^{100}}{100}\)
Sums, products, and quotients
Differentiation of sums and differences
\(F(x)=f(x)\pm g(x) \Longrightarrow f'(x)\pm g'(x)\)
Proof
\(F'(x)=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}\Longrightarrow \lim_{h\to 0}\frac{f(x+h)+g(x+h)-(f(x)+g(x))}{h}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}+\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\)
Which implies
\(F'(x)=f'(x)+g'(x)\)
Products
\(F(x)=f(x)\cdot g(x)\Longrightarrow F'(x)=f'(x)g(x)+f(x)g'(x)\)
Proof
\(F'(x)=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}\Longrightarrow \lim_{h\to 0}\frac{f(x+h)\cdot g(x+h)-f(x)g(x)}{h}\)
We now add and subtract \(f(x)g(x+h)\) to the numerator
\(F'(x)=\lim_{h\to 0}\frac{f(x+h)\cdot g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}\Longrightarrow\lim_{h\to 0}\frac{[f(x+h)-f(x)]g(x+h)+f(x)[g(x+h)-g(x)]}{h}\)
Rearranging yields
\(F'(x)=\lim_{h\to 0}g(x+h)\frac{f(x+h)-f(x)}{h}+f(x)\frac{g(x+h)-g(x)}{h}\)
This implies
\(F'(x)=g(x)f'(x)+f(x)g'(x)\)
Example
Find the derivative of
\(h(x)=(x^3-x)(5x^4+x^2)\)
Quotients
\(F(x)=\frac{f(x)}{g(x)}\Longrightarrow F'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\)
Examples
- \(F(x)=\frac{3x-5}{x-2}\)
- Find the derivative for the average cost function \(AC=\frac{C(x)}{x}\)