Derivatives of Polynomials, Trigonometric and Logarithmic Functions
The derivative is a central tool in differential calculus, with fundamental applications in science, engineering, and economics. This article offers a progressive guide to mastering the differentiation of functions, from polynomials to trigonometric and logarithmic functions. Through demonstrations and concrete examples, it aims to foster an understanding of both the application of the rules and their underlying principles.
Learning Objectives
- Understand the general concept of derivative and its fundamental properties.
- Apply the formal definition of derivative to compute basic derivatives.
- Demonstrate by means of limits the derivative of constant functions and the identity function.
- Obtain the rules for differentiating trigonometric functions starting from the fundamental derivatives of sine and cosine.
- Compute derivatives of composite trigonometric functions using algebraic rules.
- Formally demonstrate the derivative of the natural logarithm using limits.
TABLE OF CONTENTS:
Derivative of Algebraic Functions
Derivatives of Transcendental Functions
So far, we have only reviewed what the derivative is and some of its algebraic properties, but we have not addressed how to actually compute it. Here we will resolve that issue by showing each differentiation technique and how it is derived in each case.
Derivative of Algebraic Functions
Constant Function
If f(x) = c, where c is any real constant, then we have:
\displaystyle \frac{df(x)}{dx} =\frac{d}{dx}c = 0
PROOF: This proof is actually done in a single step:
\begin{array}{rll} (1) &\displaystyle \dfrac{d}{dx}c &=\displaystyle \lim_{\Delta x \to 0} \dfrac{c - c}{\Delta x} \quad \text{; Definition of derivative for $f(x)=c$} \\ \\ & &=\displaystyle \lim_{\Delta x \to 0} \dfrac{0}{\Delta x} = 0 \end{array}
Identity Function
\displaystyle \frac{df(x)}{dx} =\frac{dx}{dx}=1
PROOF: Very similar to the previous one, also derived in a single step:
\begin{array}{rll} (1) & \dfrac{d}{dx}x &= \displaystyle \lim_{\Delta x\to 0} \dfrac{(x+\Delta x) - x}{\Delta x} \quad \text{; Definition of derivative for $f(x) = x$}\\ \\ & &=\displaystyle \lim_{\Delta x\to 0} \dfrac{\Delta x}{\Delta x} = 1 \end{array}
Natural Powers
If f(x) = x^n, where n is any natural number, then we have:
\displaystyle \frac{df(x)}{dx} =\frac{dx^n}{dx} =nx^{n-1}
PROOF: To prove this theorem, we will need to use the Binomial Theorem
\begin{array}{rll} (1) &\displaystyle \dfrac{d}{dx}x^n = \lim_{\Delta x \to 0} \frac{(x+\Delta x)^n -x^n}{\Delta x} &\text{ ; Definition of limit for $f(x)= x^n$} \\ \\ & \displaystyle \phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \dfrac{\displaystyle \left[\sum_{k=0}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k} \right] - x^n}{\Delta x} & \text{; Binomial Theorem applied to $(1)$} \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \dfrac{\displaystyle x^n + \left[\sum_{k=1}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k} \right] - x^n}{\Delta x} & \text{; Isolating the first term of the sum} \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \dfrac{\displaystyle \left[\sum_{k=1}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k} \right]}{\Delta x} & \text{; Canceling like terms} \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \displaystyle \left[\sum_{k=1}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k-1} \right] & \\ \\ & \displaystyle\phantom{\displaystyle \dfrac{d}{dx}x^n} =\lim_{\Delta x \to 0} \displaystyle \left[ {{n}\choose{1}} x^{n-1}(\Delta x)^{0} + \sum_{k=2}^n {{n}\choose{k}} x^{n-k}(\Delta x)^{k-1} \right] & \text{; Extracting the first term from the sum} \\ \\ & \displaystyle \color{blue} {\displaystyle \dfrac{d}{dx}x^n} = n x^{n-1} & \color{black} \end{array}
Integer Powers
The proof just reviewed only supports the case in which the exponents are natural numbers, but it can be extended to arbitrary integers. If a\in \mathbb{Z}, then we have:
\displaystyle \frac{dx^a}{dx} = ax^{a-1}
We already know this works for positive integers; we only need to examine what happens when we consider negative powers. It is therefore sufficient to prove that:
\displaystyle \frac{dx^{-n}}{dx} = {-n}x^{-n-1}
PROOF: To do this, it is enough to consider the derivative of the quotient:
\begin{array}{rll} (1) & \dfrac{d}{dx}x^{-n} &= \dfrac{d}{dx} \left( \dfrac{1}{x^n}\right) \\ \\ & &= \dfrac{0 \cdot nx^{n-1} - nx^{n-1} \cdot 1}{x^{2n}}\\ \\ & &= -nx^{n-1-2n} \\ \\ & &= -nx^{-n-1} \end{array}
Derivatives of Transcendental Functions
Trigonometric Functions
These include the following differentiation rules:
| \displaystyle \frac{d}{dx}\sin(x) = \cos(x) | \displaystyle \frac{d}{dx}\sec(x) = \sec(x)\tan(x) |
| \displaystyle \frac{d}{dx}\cos(x) = -\sin(x) | \displaystyle \frac{d}{dx}\csc(x) = -\csc(x)\cot(x) |
| \displaystyle \frac{d}{dx}\tan(x) = \sec^2(x) | \displaystyle \frac{d}{dx}\cot(x) = -\csc^2(x) |
To obtain each of these rules, the best approach is to begin with the derivatives of sine and cosine; then, using the results along with the algebra of derivatives, derive the formulas for the other trigonometric functions.
Proof of the Derivative of Sine
\begin{array}{rll} (1) &\dfrac{d}{dx}\sin(x) = \displaystyle \lim_{\Delta x \to 0} \dfrac{\sin(x+\Delta x) - \sin(x)}{\Delta x} & \text{; Definition of derivative of sine} \\ \\ &\phantom{\dfrac{d}{dx}\sin(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{\sin(x)\cos(\Delta x) + \sin(\Delta x)\cos(x) - \sin(x)}{\Delta x} & \\ \\ &\phantom{\dfrac{d}{dx}\sin(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{ \sin(x)\left[\cos(\Delta x) -1\right] + \sin(\Delta x)\cos(x) }{\Delta x} & \\ \\ &\phantom{\dfrac{d}{dx}\sin(x)} = \displaystyle \sin(x)\lim_{\Delta x \to 0} \left[\dfrac{\cos(\Delta x) - 1}{\Delta x} \right] + \cos(x) \lim_{\Delta x \to 0} \left[ \dfrac{\sin(\Delta x)}{\Delta x} \right] & \\ \\ (2)&\displaystyle \lim_{\Delta x\to 0} \dfrac{\sin(\Delta x)}{\Delta x} = 1 & \text{; By the Squeeze Theorem}\\ \\ (3)&\displaystyle \lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x} = \lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x} \cdot \dfrac{\cos(\Delta x) + 1}{\cos(\Delta x) + 1} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} = \lim_{\Delta x\to 0} \dfrac{\cos^2(\Delta x) - 1}{\Delta x (\cos(\Delta x) + 1)} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} = \lim_{\Delta x\to 0} \dfrac{-\sin^2(\Delta x)}{\Delta x (\cos(\Delta x) + 1)} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} =- \lim_{\Delta x\to 0} \dfrac{\sin(\Delta x)}{\Delta x} \cdot \lim_{\Delta x\to 0} \dfrac{\sin(\Delta x)}{\cos(\Delta x) + 1} & \\ \\ &\displaystyle\phantom{\lim_{\Delta x\to 0} \dfrac{\cos(\Delta x) - 1}{\Delta x}} =- (1)\cdot(0) = 0 \\ \\ (4) &\color{blue}\dfrac{d}{dx}\sin(x) = \cos(x) \color{black} & \text{; From (1,2,3)} \end{array}
Proof of the Derivative of Cosine
\begin{array}{rll} (1) & \dfrac{d}{dx}\cos(x) = \displaystyle \lim_{\Delta x \to 0} \dfrac{\cos(x + \Delta x) - \cos(x)}{\Delta x} & \text{; Definition of the derivative of cosine} \\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{\cos(x)\cos(\Delta x) - \sin(x)\sin(\Delta x) - \cos(x)}{\Delta x} \\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \displaystyle \lim_{\Delta x \to 0} \dfrac{\cos(x) [ \cos(\Delta x) - 1] - \sin(x)\sin(\Delta x)}{\Delta x}\\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \cos(x) \displaystyle \lim_{\Delta x \to 0} \dfrac{ [ \cos(\Delta x) - 1]}{\Delta x} - \sin(x) \lim_{\Delta x \to 0} \dfrac{\sin(\Delta x)}{\Delta x}\\ \\ & \phantom{\dfrac{d}{dx}\cos(x)} = \cos(x) \cdot(0) - \sin(x)\cdot (1)\\ \\ &\color{blue}\dfrac{d}{dx}\cos(x) = - \sin(x) \color{black} \end{array}
Derivatives of Tangent, Secant, Cosecant, and Cotangent
With the results for sine and cosine already obtained, deriving the formulas for the remaining trigonometric functions is now straightforward.
\begin{array}{rl} \dfrac{d}{dx}\tan(x) &= \dfrac{d}{dx} \left( \dfrac{\sin(x)}{\cos(x)} \right) = \dfrac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \dfrac{1}{\cos^2(x)} = \color{blue}\sec^2(x) \color{black}\\ \\ \dfrac{d}{dx}\sec(x) &= \dfrac{d}{dx}\left(\dfrac{1}{\cos(x)} \right) = \dfrac{\sin(x)}{\cos^2(x)} =\color{blue}\sec(x)\tan(x) \color{black}\\ \\ \dfrac{d}{dx}\csc(x) &= \dfrac{d}{dx}\left(\dfrac{1}{\cos(x)}\right) = -\dfrac{cos(x)}{\sin^2(x)} =\color{blue} - \csc(x)\cot(x)\color{black}\\ \\ \dfrac{d}{dx} \cot(x) &= \dfrac{d}{dx} \left(\dfrac{\cos(x)}{\sin(x)}\right) = \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} = -\dfrac{1}{\sin^2(x)} =\color{blue} -\csc^2(x)\color{black} \end{array}
Derivative of Logarithmic Functions
The derivative of the natural logarithm is given by:
\displaystyle \frac{d}{dx}\ln(x) = \frac{1}{x}
PROOF: Reasoning from the definition of the derivative, we obtain the following argument:
\begin{array}{rll} (1) & \dfrac{d}{dx} \ln(x) = \displaystyle \lim_{\Delta x \to 0} \left [\dfrac{\ln(x+\Delta x) - \ln(x)}{\Delta x} \right] &\text{; Definition of the derivative for Natural Logarithm} \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \dfrac{1}{\Delta x} \ln \left( \dfrac{x+\Delta x}{x} \right) \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \ln \left( \dfrac{x+\Delta x}{x} \right)^{\frac{1}{\Delta x} } \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \ln \left( \dfrac{x+\Delta x}{x} \right)^{\frac{1}{\color{red}x\color{black}} \frac{\color{red}x\color{black}}{\Delta x} } \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} = \displaystyle \lim_{\Delta x \to 0} \left[ \dfrac{1}{x} \ln \left( 1 + \dfrac{\Delta x}{x} \right)^{ \frac{x}{\Delta x} } \right] \\ \\ & \phantom{\dfrac{d}{dx} \ln(x)} =\dfrac{1}{x} \ln \displaystyle \left[ \lim_{\Delta x \to 0} \left( 1 + \dfrac{\Delta x}{x} \right)^{ \frac{x}{\Delta x} } \right] \\ \\ (2) & n=\dfrac{x}{\Delta x} & \text{; substitution} \\ \\ (3) & (\Delta x \to 0^+) \longrightarrow (n\to +\infty) \\ \\ (4) & \dfrac{d}{dx} \ln(x) = \dfrac{1}{x} \ln\left[ \displaystyle \lim_{n \to +\infty} \left(1 + \dfrac{1}{n} \right)^n \right] = \dfrac{1}{x} \ln(e) = \color{blue}\dfrac{1}{x} \color{black} & \text{; From (1,2,3)} \end{array}
With this, we have step by step covered the fundamental derivatives that every student must master: from basic algebraic functions to the main transcendental functions such as trigonometric functions and the natural logarithm. By mastering these proofs, you will be able to apply the rules of differentiation and understand their origin and formal justification. This knowledge forms the foundation for confidently tackling more complex problems that require a precise analysis of change.