Introduction to Derivatives (First Principle)

1. Geometric Interpretation of the Derivative

The derivative of a function at a point represents the instantaneous rate of change of the function at that point. Geometrically, this is equivalent to the slope of the tangent line to the graph of the function at that point.

Consider a function $y = f(x)$. Let $P(a, f(a))$ be a point on the curve. Let's take another point $Q(a+h, f(a+h))$ very close to $P$. The slope of the secant line connecting $P$ and $Q$ is given by: $$m_{PQ} = \frac{\text{change in y}}{\text{change in x}} = \frac{f(a+h) - f(a)}{(a+h) - a} = \frac{f(a+h) - f(a)}{h}$$

As we make $h$ smaller and smaller (i.e., as $h \to 0$), the point $Q$ gets closer to $P$. The secant line $PQ$ rotates to become the tangent line at point $P$. The slope of this tangent line is the limit of the slope of the secant line as $h \to 0$.

2. The Derivative from First Principles

This limiting value is called the derivative of the function $f(x)$ at the point $x=a$. It is denoted by $f'(a)$.

Definition (First Principle of Derivatives): The derivative of a function $f(x)$ with respect to $x$ is the function $f'(x)$ given by: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ provided this limit exists.

This formula is also known as the delta method or the ab-initio method. If the limit exists, the function is said to be differentiable or derivable at that point.

Other notations for the derivative of $y=f(x)$ include:

• $\frac{dy}{dx}$
• $y'$
• $\frac{d}{dx}f(x)$
• $D(f(x))$

3. Physical Interpretation

If $s(t)$ represents the position of a particle at time $t$, then its derivative, $s'(t)$ or $\frac{ds}{dt}$, represents the instantaneous velocity of the particle at time $t$.

Solved Examples

Example 1: Derivative of a Constant

Question: Find the derivative of $f(x) = c$ (where c is a constant) using the first principle.

Solution: We have $f(x) = c$. So, $f(x+h) = c$. Using the definition: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} \frac{0}{h} = 0$$ Thus, the derivative of any constant function is 0.

Example 2: Derivative of $f(x) = x^2$

Question: Find the derivative of $f(x) = x^2$ using the first principle.

Solution: We have $f(x) = x^2$, so $f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$. Using the definition: $$f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}$$ Factor out $h$ from the numerator: $$f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h} = \lim_{h \to 0} (2x + h) = 2x + 0 = 2x$$ So, $\frac{d}{dx}(x^2) = 2x$.

Example 3: Derivative of $f(x) = \sqrt{x}$

Question: Find the derivative of $f(x) = \sqrt{x}$ using the first principle.

Solution: We have $f(x) = \sqrt{x}$, so $f(x+h) = \sqrt{x+h}$. $$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$ This is a $\frac{0}{0}$ form. We rationalize the numerator: $$f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$ $$= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$$

Example 4: Derivative of $f(x) = \sin x$

Question: Find the derivative of $f(x) = \sin x$ using the first principle.

Solution: We have $f(x) = \sin x$, so $f(x+h) = \sin(x+h)$. $$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$$ Using the trigonometric identity $\sin C - \sin D = 2\cos(\frac{C+D}{2})\sin(\frac{C-D}{2})$: $$f'(x) = \lim_{h \to 0} \frac{2\cos(\frac{x+h+x}{2})\sin(\frac{x+h-x}{2})}{h} = \lim_{h \to 0} \frac{2\cos(x+h/2)\sin(h/2)}{h}$$ Rearrange to use standard limits: $$= \lim_{h \to 0} \cos(x+h/2) \cdot \frac{\sin(h/2)}{h/2}$$ As $h \to 0$, $h/2 \to 0$. So, $\lim_{h \to 0} \frac{\sin(h/2)}{h/2} = 1$. $$f'(x) = \lim_{h \to 0} \cos(x+h/2) \cdot 1 = \cos(x+0) = \cos x$$