To determine if a function is differentiable on an interval, the fundamental requirement is that the derivative of the function must exist for every single point within that interval. This means that at every point 'a' in the given interval 'I', the derivative f′(a) must be well-defined.
What Does Differentiability on an Interval Mean?
A function $f$ is considered differentiable on an interval $I$ if, for every point $a$ belonging to $I$, its derivative, denoted as $f'(a)$, exists. The existence of the derivative at a point implies several crucial characteristics about the function's behavior at that point.
Understanding the Derivative's Existence
For the derivative $f'(a)$ to exist at a specific point $a$, the function must meet certain criteria:
- Continuity: The function must be continuous at point $a$. If there's a break, hole, or jump in the graph, the function cannot be differentiated at that point. While differentiability implies continuity, the reverse is not always true (a continuous function is not necessarily differentiable).
- Smoothness: The graph of the function must be "smooth" at point $a$, meaning it cannot have sharp corners (cusps) or kinks. At a sharp corner, the slope changes abruptly, making it impossible to define a unique tangent line.
- No Vertical Tangents: The function's graph cannot have a vertical tangent line at point $a$. A vertical tangent implies an infinite slope, meaning the derivative is undefined at that point.
Key Conditions for Differentiability
The presence or absence of these features helps in determining differentiability.
Condition for Differentiability | Description | Example of Violation |
---|---|---|
Continuity | The function must have no breaks, holes, or jumps in its graph. | $f(x) = \frac{1}{x}$ at $x=0$ |
Smoothness | The graph must not have any sharp corners or cusps. The slope must transition smoothly. | $f(x) = |
No Vertical Tangent | The tangent line at any point must not be vertical (i.e., the slope must not be infinite). | $f(x) = \sqrt[3]{x}$ at $x=0$ |
Derivative Exists at Every Point | The limit defining the derivative must exist from both sides and be equal at every point in the interval. | (Applies to the core definition for the entire interval) |
How to Systematically Check Differentiability
Determining differentiability on an interval often involves a combination of analytical and graphical methods.
Step-by-Step Analytical Approach (Especially for Piecewise Functions)
For functions defined by a single rule (e.g., polynomials, exponentials), differentiability is usually straightforward. However, for piecewise functions or functions with specific domain restrictions, a systematic approach is crucial.
- Check Differentiability of Each Piece:
- Verify that each component function is differentiable on its respective open sub-interval. For example, polynomials, exponential functions ($e^x$), sine ($\sin x$), and cosine ($\cos x$) are differentiable everywhere. Rational functions are differentiable on their domains (where the denominator is not zero).
- Examine Critical Points (Boundary Points/Transition Points):
- These are the points where the function's definition changes.
- Check for Continuity: First, ensure the function is continuous at the critical point. If it's not continuous, it cannot be differentiable.
- Calculate $f(a)$, $\lim{x \to a^-} f(x)$, and $\lim{x \to a^+} f(x)$. All three must be equal.
- Check for Smoothness (Derivative Existence): If continuous, calculate the left-hand derivative ($f'(a^-)$) and the right-hand derivative ($f'(a^+)$) at the critical point.
- The function is differentiable at the critical point only if $f'(a^-) = f'(a^+)$. If they are unequal, there is a sharp corner or cusp.
Graphical Inspection
A quick visual check can often reveal potential issues:
- Smooth Curve: Look for a continuous curve without any breaks.
- No Sharp Turns: Avoid points where the graph forms a "V" shape or a cusp.
- No Vertical Segments: Ensure there are no points where the tangent line would be vertical.
Properties of Common Functions
- Polynomial Functions: Differentiable everywhere on $(-\infty, \infty)$.
- Rational Functions: Differentiable on their domain (where the denominator is not zero).
- Exponential Functions ($e^x, a^x$): Differentiable everywhere on $(-\infty, \infty)$.
- Logarithmic Functions ($\ln x, \log_a x$): Differentiable on their domain $(0, \infty)$.
- Trigonometric Functions ($\sin x, \cos x$): Differentiable everywhere on $(-\infty, \infty)$.
- Absolute Value Function ($|x|$): Differentiable everywhere except at $x=0$ (due to a sharp corner).
- Root Functions ($\sqrt{x}, \sqrt[3]{x}$): $\sqrt{x}$ is differentiable on $(0, \infty)$, not at $x=0$ (vertical tangent). $\sqrt[3]{x}$ is differentiable everywhere except $x=0$ (vertical tangent).
Examples Illustrating Differentiability
Here are a few examples to clarify the concept:
-
Function Differentiable Everywhere:
- $f(x) = x^3 - 2x + 1$
- Since this is a polynomial, its derivative $f'(x) = 3x^2 - 2$ exists for all real numbers. Thus, $f(x)$ is differentiable on $(-\infty, \infty)$.
-
Function Not Differentiable at a Point:
- $g(x) = |x|$
- This function is continuous everywhere. However, its graph has a sharp corner at $x=0$.
- The left-hand derivative at $x=0$ is $-1$, and the right-hand derivative at $x=0$ is $1$. Since $f'(0^-) \neq f'(0^+)$, $g(x)$ is not differentiable at $x=0$. It is differentiable on $(-\infty, 0) \cup (0, \infty)$.
-
Piecewise Function Example:
- Consider the function:
$h(x) = \begin{cases} x^2 & \text{if } x \le 1 \ 2x - 1 & \text{if } x > 1 \end{cases}$ - Step 1: Check each piece:
- For $x \le 1$, $x^2$ is differentiable. $h'(x) = 2x$.
- For $x > 1$, $2x-1$ is differentiable. $h'(x) = 2$.
- Step 2: Check at the transition point ($x=1$):
- Continuity:
- $h(1) = 1^2 = 1$
- $\lim{x \to 1^-} h(x) = \lim{x \to 1^-} x^2 = 1$
- $\lim{x \to 1^+} h(x) = \lim{x \to 1^+} (2x - 1) = 2(1) - 1 = 1$
- Since all three are equal, $h(x)$ is continuous at $x=1$.
- Differentiability (Smoothness):
- Left-hand derivative: $\lim{x \to 1^-} h'(x) = \lim{x \to 1^-} 2x = 2(1) = 2$
- Right-hand derivative: $\lim{x \to 1^+} h'(x) = \lim{x \to 1^+} 2 = 2$
- Since the left-hand derivative equals the right-hand derivative ($2=2$), $h(x)$ is differentiable at $x=1$.
- Continuity:
- Conclusion: $h(x)$ is differentiable on $(-\infty, \infty)$.
- Consider the function:
Practical Considerations and Tips
- Always remember that continuity is a prerequisite for differentiability. If a function is not continuous at a point, it cannot be differentiable at that point.
- Most elementary functions (polynomials, exponentials, sines, cosines) are differentiable on their entire domains.
- Pay close attention to points where the function's definition changes, or where a denominator might be zero, or where a term under a square root becomes negative. These are common locations for non-differentiability.
- When faced with a complex function, breaking it down into simpler components and applying the rules of differentiation (chain rule, product rule, etc.) can help determine if the derivative exists.
For deeper insights into specific function types and their differentiability, consulting reputable calculus textbooks or online educational resources, such as those found on university mathematics department websites, can be highly beneficial.