Definition of the Derivative

The derivative is one of the most fundamental concepts in calculus. It measures the instantaneous rate of change of a quantity. Geometrically, the derivative of a function at a specific point is the slope of the tangent line to the function's curve at that point.

The Limit Definition of the Derivative

The formal definition of the derivative is based on the limit of the difference quotient. For a function $f(x)$, its derivative $f'(x)$ is defined as:

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

Here, $\frac{f(x+h) - f(x)}{h}$ is the difference quotient, which represents the average rate of change over a small interval $h$. By taking the limit as $h$ approaches $0$, we find the exact instantaneous rate of change at the point $x$.

Differentiability and Continuity

For a function to have a derivative at a point (meaning it is differentiable), it must be continuous at that point.

However, continuity alone does not guarantee differentiability. If a graph has a sharp corner or cusp (like $f(x) = |x|$ at $x=0$), or a vertical tangent line, it is continuous but not differentiable at those specific points.

Example Problems

Example 1: Using the Limit Definition

Problem: Use the limit definition to find $f'(x)$ for $f(x) = x^2 + 3x$.

Solution:

1. Set up the difference quotient by finding $f(x+h)$: $f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$
2. Subtract $f(x)$: $f(x+h) - f(x) = (x^2 + 2xh + h^2 + 3x + 3h) - (x^2 + 3x) = 2xh + h^2 + 3h$
3. Divide by $h$: $\frac{2xh + h^2 + 3h}{h} = 2x + h + 3$
4. Take the limit as $h \to 0$: $f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 0 + 3 = 2x + 3$

Example 2: Finding the Tangent Line

Problem: Find the equation of the tangent line to $y = x^3$ at $x = 2$.

Solution:

1. Find the derivative to get the slope formula (using the power rule): $f'(x) = 3x^2$
2. Find the slope at $x = 2$: $m = f'(2) = 3(2)^2 = 12$
3. Find the $y$-coordinate at $x = 2$: $y = (2)^3 = 8$ The point of tangency is $(2, 8)$.
4. Use the point-slope formula ($y - y_1 = m(x - x_1)$) to write the equation: $y - 8 = 12(x - 2)$ $y = 12x - 24 + 8$ $y = 12x - 16$