When Is a Function Differentiable? Revision Notes for VCE SSCE Mathematical Methods

When Is a Function Differentiable?

What does it mean for a function to be differentiable?

We say a function $f$ is differentiable at a point $x = a$ when the following limit exists:

\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

This limit represents the derivative of the function at that point. If this limit exists, we can find the gradient of the tangent line to the function at $x = a$ .

Most functions you'll work with are differentiable throughout their domains. But not all functions have this property.

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For instance, the absolute value function $f(x) = |x|$ is not differentiable at $x = 0$ because the gradient is $-1$ to the left of zero and $1$ to the right of zero. The limit from the left and right don't match, so the derivative doesn't exist at that point.

Piecewise functions and differentiability

Some piecewise-defined functions are differentiable everywhere, while others are not. The key factor is whether the pieces join smoothly at the boundary points.

When two pieces of a piecewise function meet at a point:

If they join smoothly (no sharp corner), the function is differentiable at that point
If they form a sharp corner or discontinuity, the function is not differentiable at that point

Let's look at some examples to understand this better.

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Worked Example: Smooth Join (Differentiable Everywhere)

For the function with the following rule, find $f'(x)$ and sketch the graph of $y = f'(x)$ :

f(x) = \begin{cases} x^2 + 2x + 1 & \text{if } x \geq 0 \\ 2x + 1 & \text{if } x < 0 \end{cases}

Solution:

To find the derivative, we differentiate each piece separately:

For $x \geq 0$ : $\frac{d}{dx}(x^2 + 2x + 1) = 2x + 2$

For $x < 0$ : $\frac{d}{dx}(2x + 1) = 2$

Therefore:

f'(x) = \begin{cases} 2x + 2 & \text{if } x \geq 0 \\ 2 & \text{if } x < 0 \end{cases}

Importantly, $f'(0)$ is defined and equals $2$ . This is because both pieces give the same derivative value at $x = 0$ . The two sections of the original function join smoothly at the point $(0, 1)$ .

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Worked Example: Sharp Corner (Not Differentiable at Join)

For the function with the following rule, state where the derivative is defined, find $f'(x)$ for this set of values, and sketch the graph of $y = f'(x)$ :

f(x) = \begin{cases} x^2 + 2x + 1 & \text{if } x \geq 0 \\ x + 1 & \text{if } x < 0 \end{cases}

Solution:

Again, we differentiate each piece:

For $x > 0$ : $\frac{d}{dx}(x^2 + 2x + 1) = 2x + 2$

For $x < 0$ : $\frac{d}{dx}(x + 1) = 1$

Therefore:

f'(x) = \begin{cases} 2x + 2 & \text{if } x > 0 \\ 1 & \text{if } x < 0 \end{cases}

Here's the crucial difference: $f'(0)$ is not defined because the limits from the left and right are not equal. From the left, the gradient approaches $1$ , while from the right it approaches $2$ .

The function $f$ is differentiable for all real numbers except zero, which we write as $\mathbb{R} \backslash \{0\}$ .

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Worked Example: Continuous but Not Differentiable

For the function with rule $f(x) = x^{\frac{1}{3}}$ , state when the derivative is defined and sketch the graph of the derivative function.

Solution:

Using the power rule for differentiation:

f'(x) = \frac{1}{3}x^{-\frac{2}{3}}

The derivative is not defined at $x = 0$ because we cannot divide by zero (the power $-\frac{2}{3}$ creates a zero in the denominator).

We can verify this using first principles. The derivative at $x = 0$ would be:

\frac{f(0+h) - f(0)}{h} = \frac{h^{\frac{1}{3}} - 0}{h} = \frac{h^{\frac{1}{3}}}{h} = h^{-\frac{2}{3}}

As $h \to 0$ , we have $h^{-\frac{2}{3}} \to \infty$ . Since this limit doesn't exist, $f'(0)$ is not defined.

This example shows something important: the function $f(x) = x^{\frac{1}{3}}$ is continuous everywhere (you can draw it without lifting your pen), but it's not differentiable at $x = 0$ (the tangent line is vertical at that point).

The relationship between differentiability and continuity

chatImportant

If a function is differentiable at a point, then it must also be continuous at that point.

However, the reverse is not true! A function can be continuous at a point but not differentiable there. The cube root function example above demonstrates this perfectly: it is continuous at $x = 0$ , but not differentiable there.

Think of it this way:

Differentiability is a stronger condition than continuity
Differentiable $\implies$ continuous (always true)
Continuous $\not\implies$ differentiable (not always true)

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Key Points to Remember:

A function is differentiable at $x = a$ if the limit $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists
For piecewise functions, check if the pieces join smoothly - if they form a sharp corner, the function is not differentiable at that point
Differentiability implies continuity, but continuity does not imply differentiability
A function can be continuous everywhere but still have points where it's not differentiable (like $f(x) = x^{\frac{1}{3}}$ at $x = 0$ )
When finding derivatives of piecewise functions, check the derivative from both sides at boundary points to determine if the function is differentiable there

When Is a Function Differentiable? (VCE SSCE Mathematical Methods): Revision Notes