Derivative from First Principles (Ab Initio Approach)

For a function \( y = f(x) \), the derivative is given by

\[ \frac{dy}{dx} = \frac{d}{dx} f(x) = f'(x). \]

Using the limit definition,

\[ f'(x) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}. \]

When we find the derivative of a function \( y = f(x) \), written as \( \frac{dy}{dx} \), we say that we are differentiating \( y \) with respect to \( x \). This process of computing the derivative is called differentiation.

Let us apply on some of the elementary functions.

I. For the function \( y = x^a \) where \( a \in \mathbb{R} \), the derivative is given by

\[ \frac{dy}{dx} = ax^{a-1}. \]

To establish this result rigorously for rational values of \( a \), we use the first principles definition of the derivative:

\[ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}. \]

Substituting \( f(x) = x^a \),

\[ \lim_{\Delta x \to 0} \frac{(x + \Delta x)^a - x^a}{\Delta x}. \]

Using the binomial expansion for small \( \Delta x \) (valid when \( a \in \mathbb{Q} \)),

\[ (x + \Delta x)^a = x^a \left( 1 + \frac{a \Delta x}{x} + \frac{a (a-1)}{2} \left(\frac{\Delta x}{x}\right)^2 + \dots \right). \]

Thus,

\[ (x + \Delta x)^a - x^a = x^a \left( 1 + \frac{a \Delta x}{x} + \frac{a (a-1)}{2} \left(\frac{\Delta x}{x}\right)^2 + \dots \right) - x^a. \]

Simplifying,

\[ (x + \Delta x)^a - x^a = x^a \left( \frac{a \Delta x}{x} + \frac{a (a-1)}{2} \left(\frac{\Delta x}{x}\right)^2 + \dots \right). \]

Dividing by \( \Delta x \),

\[ \frac{(x + \Delta x)^a - x^a}{\Delta x} = x^a \left( \frac{a}{x} + \frac{a (a-1)}{2} \frac{\Delta x}{x^2} + \dots \right). \]

Taking the limit as \( \Delta x \to 0 \), higher-order terms vanish, leaving

\[ \lim_{\Delta x \to 0} \frac{(x + \Delta x)^a - x^a}{\Delta x} = a x^{a-1}. \]

Thus, for rational values of \( a \),

\[ \frac{d}{dx} x^a = a x^{a-1}. \]

This result extends to irrational values of \( a \) using the continuity of the derivative function and the density of rationals in the real numbers.

II. Let us find the derivative of \( \sin x \) using the first principles definition. The derivative is given by

\[ \frac{d}{dx} \sin x = \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin x}{\Delta x}. \]

Using the sum-to-product identity,

\[ \sin(x + \Delta x) - \sin x = 2 \cos \left( x + \frac{\Delta x}{2} \right) \sin \left( \frac{\Delta x}{2} \right), \]

we substitute this into the limit:

\[ \lim_{\Delta x \to 0} \frac{2 \cos \left( x + \frac{\Delta x}{2} \right) \sin \left( \frac{\Delta x}{2} \right)}{\Delta x} = \lim_{\Delta x \to 0} 2 \cos \left( x + \frac{\Delta x}{2} \right) \cdot \frac{\sin \left( \frac{\Delta x}{2} \right)}{\Delta x}. \]

Rewriting the fraction,

\[ \lim_{\Delta x \to 0} 2 \cos \left( x + \frac{\Delta x}{2} \right) \cdot \frac{\sin \left( \frac{\Delta x}{2} \right)}{\frac{\Delta x}{2}} \cdot \frac{1}{2}. \]

Since

\[ \lim_{\Delta x \to 0} \frac{\sin \left( \frac{\Delta x}{2} \right)}{\frac{\Delta x}{2}} = 1, \]

we get

\[ \lim_{\Delta x \to 0} 2 \cos \left( x + \frac{\Delta x}{2} \right) \cdot 1 \cdot \frac{1}{2} = \lim_{\Delta x \to 0} \cos \left( x + \frac{\Delta x}{2} \right). \]

Taking the limit,

\[ \cos x. \]

Thus,

\[ \frac{d}{dx} \sin x = \cos x. \]

This completes the proof using the first principles definition.

III. Let us find the derivative of \( e^x \) using the first principles definition. The derivative is given by

\[ \frac{d}{dx} e^x = \lim_{\Delta x \to 0} \frac{e^{x + \Delta x} - e^x}{\Delta x}. \]

Rewriting the numerator using exponent properties,

\[ e^{x + \Delta x} - e^x = e^x e^{\Delta x} - e^x = e^x (e^{\Delta x} - 1). \]

Substituting this into the limit,

\[ \frac{d}{dx} e^x = \lim_{\Delta x \to 0} \frac{e^x (e^{\Delta x} - 1)}{\Delta x} = e^x \lim_{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x}. \]

The key limit result,

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

follows from the definition of \( e \) as the base of the natural logarithm. Applying this,

\[ \frac{d}{dx} e^x = e^x \cdot 1 = e^x. \]

Thus,

\[ \frac{d}{dx} e^x = e^x. \]

This proves that the derivative of \( e^x \) is \( e^x \), a fundamental property of the exponential function.

IV. Let us find the derivative of \( \ln x \) using the first principles definition. The derivative is given by

\[ \frac{d}{dx} \ln x = \lim_{\Delta x \to 0} \frac{\ln (x + \Delta x) - \ln x}{\Delta x}. \]

Using the logarithmic property,

\[ \ln (x + \Delta x) - \ln x = \ln \left( \frac{x + \Delta x}{x} \right) = \ln \left( 1 + \frac{\Delta x}{x} \right), \]

substituting this into the limit,

\[ \frac{d}{dx} \ln x = \lim_{\Delta x \to 0} \frac{\ln \left( 1 + \frac{\Delta x}{x} \right)}{\Delta x}. \]

Rewriting,

\[ \frac{d}{dx} \ln x = \lim_{\Delta x \to 0} \frac{\ln \left( 1 + \frac{\Delta x}{x} \right)}{\frac{\Delta x}{x}} \cdot \frac{1}{x}. \]

Using the standard limit result,

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

where \( u = \frac{\Delta x}{x} \), we obtain

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

Thus,

\[ \frac{d}{dx} \ln x = \frac{1}{x}. \]

This proves that the derivative of \( \ln x \) is

\[ \frac{1}{x}. \]

Using this process, we can prove the derivatives of many other functions. We leave this as an exercise for the reader. Below is a table of commonly used functions and their derivatives:

\[ \begin{array}{|c|c|} \hline \text{Function } f(x) & \text{Derivative } f'(x) \\ \hline c \text{ (constant)} & 0 \\ x & 1 \\ x^a \quad (a \in \mathbb{R}) & a x^{a-1} \\ e^x & e^x \\ a^x & a^x \ln a \\ \ln x & \frac{1}{x} \\ \log_a x & \frac{1}{x \ln a} \\ \sin x & \cos x \\ \cos x & -\sin x \\ \tan x & \sec^2 x \\ \sec x & \sec x \tan x \\ \csc x & -\csc x \cot x \\ \cot x & -\csc^2 x \\ \sin^{-1} x & \frac{1}{\sqrt{1 - x^2}} \\ \cos^{-1} x & -\frac{1}{\sqrt{1 - x^2}} \\ \tan^{-1} x & \frac{1}{1 + x^2} \\ \cot^{-1} x & -\frac{1}{1 + x^2} \\ \sec^{-1} x & \frac{1}{|x| \sqrt{x^2 - 1}} \\ \csc^{-1} x & -\frac{1}{|x| \sqrt{x^2 - 1}} \\ \hline \end{array} \]

We will use these formulas in calculating derivatives of more complicated fucntions.