Mathematics

Theory and applications of derivatives

Definition of derivative, differentiability, geometric interpretation, calculation of maxima and minima, optimization problems, and tangent and normal lines.

Derivative of a function at a point
The derivative of the function f(x)f(x) at a point x=ax = a is the value of the limit of the difference quotient as the increment of the variable approaches 00.
f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}
Example: Derivative of y=3x2y = 3x^2 at x=2x=2
f′(2)=lim⁡h→0f(2+h)−f(2)h=lim⁡h→03(2+h)2−3(2)2hf'(2) = \lim_{h\to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h\to 0} \frac{3(2+h)^2 - 3(2)^2}{h}
=lim⁡h→012+12h+3h2−12h=lim⁡h→0h(12+3h)h=12= \lim_{h\to 0} \frac{12 + 12h + 3h^2 - 12}{h} = \lim_{h\to 0} \frac{h(12 + 3h)}{h} = 12
One-sided derivatives of a function at a point
Left-hand derivative:
f′(a−)=lim⁡h→0−f(a+h)−f(a)hf'(a^-) = \lim_{h\to 0^-} \frac{f(a+h) - f(a)}{h}
Right-hand derivative:
f′(a+)=lim⁡h→0+f(a+h)−f(a)hf'(a^+) = \lim_{h\to 0^+} \frac{f(a+h) - f(a)}{h}
Definition. Derivative function
The derivative function of a function f(x)f(x) is another function f′(x)f'(x) that associates each real number with its derivative, if it exists.
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}
A function is differentiable at x=ax = a if it is continuous and both one-sided derivatives exist and are equal.
f′(a−)=f′(a+)f'(a^-) = f'(a^+)
Just as with continuity, studying differentiability involves determining at which points the function is differentiable by analyzing its domain, and if it is a piecewise function, analyzing the points where the function changes its definition.

Example. Analyze the differentiability of the function:
f(x)={x2+2xx≤1x3+x+1x>1f(x) = \begin{cases} x^2 + 2x & x \le 1 \\ x^3 + x + 1 & x > 1 \end{cases}
f′(x)={2x+2x<13x2+1x>1f'(x) = \begin{cases} 2x + 2 & x < 1 \\ 3x^2 + 1 & x > 1 \end{cases}
Both pieces are polynomials, so they are continuous and differentiable in their respective intervals. We analyze the continuity and differentiability at x=1x=1, where the function changes its definition.

1. Continuity at x=1:
lim⁡x→1−f(x)=lim⁡x→1+f(x)=f(1)=3  ⟹  It is continuous at x=1\lim_{x\to 1^-} f(x) = \lim_{x\to 1^+} f(x) = f(1) = 3 \implies \text{It is continuous at } x=1
2. Differentiability at x=1:
f′(1−)=(2x+2)x=1=4f'(1^-) = (2x+2)_{x=1} = 4
f′(1+)=(3x2+1)x=1=4f'(1^+) = (3x^2+1)_{x=1} = 4
Both derivatives match, therefore f(x)f(x) is differentiable at x=1x=1.
The derivative of a function ff at a point corresponds to the slope of the tangent line to the curve at that point.
f′(a)=lim⁡h→0f(a+h)−f(a)h=tan⁡α\displaystyle f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h} = \tan \alpha
Geometric interpretation of the derivative as the slope of the tangent line
Also known as critical points: points where the tangent line is horizontal, meaning the derivative is zero:
f′(x)=0f'(x) = 0
The solutions, x0,x1,x2...x_0, x_1, x_2..., are the candidate points for local maxima and minima. They are checked using the second derivative test:
If f′′(x0)>0 it is a local minimumIf f′′(x0)<0 it is a local maximum\begin{aligned} \textit{If } f''(x_0) > 0 & \textit{ it is a local minimum} \\ \textit{If } f''(x_0) < 0 & \textit{ it is a local maximum} \end{aligned}
Application to optimization problems. Method:
  1. Identify the objective variable to be maximized or minimized from the problem statement.
  2. Express the objective as a function of one or more variables.
  3. Calculate the required maximum or minimum by using the previous criteria, setting the derivative f′(x)f'(x) to 0.
Example: Find two numbers whose sum is 20 and whose product is as large as possible (maximum).

1) The objective is the product PP of two numbers xx and yy:
P=x⋅y\displaystyle P = x \cdot y

2) We need to express this function in a single variable. We know that x+y=20x+y=20, so y=20−xy=20-x. Therefore:
P=x⋅(20−x)\displaystyle P = x \cdot (20-x)

3) Differentiate and set to 0:
P′(x)=20−2x=0  ⟹  x=10\displaystyle P'(x) = 20 - 2x = 0 \implies x = 10

We apply the second derivative test:
P′′(x)=−2  ⟹  P′′(10)=−2\displaystyle P''(x) = -2 \implies P''(10) = -2

Since P′′(10)<0P''(10) < 0, there is a maximum at x=10x=10, and its value is P(10)=100P(10) = 100.
Equation of the tangent line
The tangent to a function at a point has the same slope as the function itself, which is precisely the derivative of the function at that point:
m=tan⁡α=f′(a)\displaystyle m = \tan \alpha = f'(a)

Thus, the tangent line to a curve at a point is the line that passes through the point (a,f(a))(a, f(a)) and has a slope equal to f′(a)f'(a). The equation of the tangent line is given by the following expression:
Tangent line to the curve at a point
y−f(a)=f′(a)⋅(x−a)y - f(a) = f'(a) \cdot (x - a)
Equation of the normal line
The slope of the normal line to a curve at a point is the negative reciprocal of the slope of the tangent line, since they are perpendicular to each other.
mn=−1m=−1tan⁡α=−1f′(a)\displaystyle m_n = -\frac{1}{m} = -\frac{1}{\tan \alpha} = -\frac{1}{f'(a)}

Therefore, the normal line to a curve at a point x=ax=a has the following equation:
Normal line perpendicular to the tangent line
y−f(a)=−1f′(a)⋅(x−a)y - f(a) = -\frac{1}{f'(a)} \cdot (x - a)