Differentiation from first principles uses the concept of the slope of a secant line approaching the tangent line as the distance between the two points becomes infinitely small.

The gradient of the tangent line at a point $Q(x, f(x))$ is given by:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

The concept of a limit is fundamental to understanding derivatives. The limit of a function $f(x)$ as $x$ approaches a value $a$, written as $\lim_{x \to a} f(x) = L$, means that as $x$ gets arbitrarily close to $a$, the value of $f(x)$ gets arbitrarily close to $L$.

Limits are used to define the slope of a tangent line to a curve. Consider a curve $y = f(x)$ and a point $Q(x_0, f(x_0))$ on the curve. To find the tangent line at $Q$, we consider another point $P(x_0 + h, f(x_0 + h))$ near $Q$, where $h$ is a small change in $x$.

The slope of the secant line passing through points $P$ and $Q$ is:

$\frac{f(x_0 + h) - f(x_0)}{(x_0 + h) - x_0} = \frac{f(x_0 + h) - f(x_0)}{h}$

As $h$ gets smaller and smaller (approaches zero), the point $P$ moves closer and closer to $Q$. The secant line $PQ$ approaches the tangent line at $Q$. Therefore, the slope of the tangent line at $Q$ is the limit of the slope of the secant line as $h$ approaches zero:

Gradient of the curve at $Q = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$

This is the definition of the derivative of $f(x)$ at $x_0$, denoted as $f'(x_0)$.

The gradient of the curve at any point $Q(x, f(x))$ is the gradient of the tangent line at that point. The gradient of the curve is called the gradient function or derivative function, because it's derived from the original function.

The formula for differentiation from first principles of $f(x)$ is:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

This means the function is differentiated with respect to $x$. $\frac{dy}{dx}$ or $f'(x)$ is called the derivative of $y = f(x)$.

Here's a detailed explanation of differentiating $f(x) = \frac{1}{x}$ from first principles:

The formula for differentiation from first principles is:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Given $f(x) = \frac{1}{x}$, then $f(x + h) = \frac{1}{x + h}$.

Substituting into the formula:

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}$

To simplify the fraction in the numerator, find a common denominator:

$f'(x) = \lim_{h \to 0} \frac{\frac{x - (x + h)}{x(x + h)}}{h}$

Simplify the numerator:

$f'(x) = \lim_{h \to 0} \frac{\frac{-h}{x(x + h)}}{h}$

Rewrite the complex fraction as a multiplication:

$f'(x) = \lim_{h \to 0} \frac{-h}{x(x + h)} \cdot \frac{1}{h}$

Cancel the $h$ terms:

$f'(x) = \lim_{h \to 0} \frac{-1}{x(x + h)}$

Now, take the limit as $h$ approaches 0:

$f'(x) = \frac{-1}{x(x + 0)}$

$f'(x) = \frac{-1}{x^2}$

Therefore, the derivative of $f(x) = \frac{1}{x}$ is $f'(x) = -\frac{1}{x^2}$.

Example 1

Differentiate $f(x) = 7x + 6$ from first principles.

Solution:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{(7(x + h) + 6) - (7x + 6)}{h}$

$= \lim_{h \to 0} \frac{7x + 7h + 6 - 7x - 6}{h} = \lim_{h \to 0} \frac{7h}{h} = \lim_{h \to 0} 7 = 7$

Therefore, $f'(x) = 7$.

Example 2

Find $f'(x)$ from first principles if $f(x) = x^2 - 6x + 1$.

Solution:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{((x + h)^2 - 6(x + h) + 1) - (x^2 - 6x + 1)}{h}$

$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 6x - 6h + 1 - x^2 + 6x - 1}{h}$

$= \lim_{h \to 0} \frac{2xh + h^2 - 6h}{h} = \lim_{h \to 0} (2x + h - 6) = 2x - 6$

Therefore, $f'(x) = 2x - 6$.

Example 3

Find the gradient function of $f(x) = x^3 + 7$ using first principles, and evaluate the gradient at $x = 2$.

Solution:

$f'(x) = \lim_{h \to 0} \frac{((x + h)^3 + 7) - (x^3 + 7)}{h}$

$= \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 + 7 - x^3 - 7}{h}$

$= \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2$

Therefore, $f'(x) = 3x^2$. At $x = 2$, $f'(2) = 3(2)^2 = 12$.

Example 4

Differentiate $f(x) = \frac{1}{x}$ from first principles.

Solution:

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x - (x + h)}{x(x + h)}}{h}$

$= \lim_{h \to 0} \frac{-h}{hx(x + h)} = \lim_{h \to 0} \frac{-1}{x(x + h)} = \frac{-1}{x^2}$

Therefore, $f'(x) = -\frac{1}{x^2}$.