A Sufficient Condition for Differentiability

Let $ f(x) $ be a function that is continuous on the interval $[a,b]$ and differentiable on the open interval $ (a,b) $, except possibly at a single point $x_0 \in (a,b)$. If \[ \lim_{x \to x_0^-} f'(x)=\lim_{x \to x_0^+} f'(x)=l, \] then $f$ is differentiable at $x_0$, and \[ f'(x_0)=l. \]

In simple terms, if the derivative approaches the same value from both sides of a point and the function is continuous there, then the function is differentiable at that point, and its derivative equals that common value.

This criterion provides a sufficient condition for differentiability, but it is not a necessary one. A function may still be differentiable even if this criterion cannot be applied.

For that reason, it is an extremely useful result in calculus, especially when computing the difference quotient directly would be difficult.

Why is this theorem useful? To determine whether a function is differentiable at a point, the standard approach is to compute the limit of the difference quotient: \[ f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}. \] In many cases, however, this limit is tedious or difficult to evaluate. If the derivative is already known on both sides of the point, this criterion offers a much simpler alternative. As long as the function is continuous and the left-hand and right-hand limits of the derivative both exist and are equal, you can immediately conclude that the function is differentiable at that point.

A Practical Example
Proof

A Practical Example

Consider the function

$$ f(x)=x|x|. $$

We want to determine whether it is differentiable at $x_0=0$ using the sufficient condition for differentiability.

First, rewrite the function without the absolute value:

$$ f(x)= \begin{cases} -x^2 & \text{if } x<0 \\ x^2 & \text{if } x\ge0 \end{cases} $$

Check that the function is continuous at $x=0$:

$$ \lim_{x\to0^-}(-x^2)=\lim_{x\to0^+}x^2=0=f(0). $$

Next, compute the derivative away from the point $0$.

For $x<0$

$$ f'(x)=-2x $$

For $x>0$

$$ f'(x)=2x. $$

Therefore, the derivative exists everywhere except, possibly, at $x=0$.

Now evaluate the one-sided limits of the derivative.

From the left,

$$ \lim_{x\to0^-}(-2x)=0. $$

From the right,

$$ \lim_{x\to0^+}2x=0. $$

Since both limits exist and are equal,

$$ \lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)=0. $$

The function is continuous at $0$, differentiable for $x\neq0$, and the left-hand and right-hand limits of its derivative agree. Therefore, the criterion tells us that

$$ f'(0)=0. $$

Notice that we never had to compute the difference quotient

$$ \lim_{h\to0}\frac{f(h)-f(0)}{h}. $$

Instead, differentiability follows directly from the behavior of the derivative near the point.

Note. This example is intentionally simple. The real advantage of the criterion becomes apparent when the derivative has a much more complicated expression and computing the difference quotient directly would require significantly more work.

Example 2

Consider the function

$$ f(x)= \begin{cases} \sin x+x^2 & \text{if } x<0 \\ x & \text{if } x\ge0 \end{cases} $$

Again, we want to determine whether the function is differentiable at $x_0=0$.

Start by checking continuity.

$$ \lim_{x\to0^-}(\sin x+x^2)=\lim_{x\to0^+}x=0=f(0). $$

Now compute the derivative for all points other than zero.

For $x<0$

$$ f'(x)=\cos x+2x $$

For $x>0$

$$ f'(x)=1. $$

Thus, the derivative exists everywhere except, possibly, at $x=0$.

Next, evaluate the one-sided limits.

From the left,

$$ \lim_{x\to0^-}(\cos x+2x)=1. $$

From the right,

$$ \lim_{x\to0^+}1=1. $$

Since

$$ \lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)=1, $$

and the function is continuous at $0$, the sufficient condition for differentiability gives

$$ f'(0)=1. $$

Once again, there is no need to evaluate the difference quotient directly:

$$ \lim_{h\to0}\frac{f(h)-f(0)}{h}. $$

The conclusion follows immediately from the behavior of the derivative on either side of the point.

Proof

Assume that $f(x)$ is continuous on the interval $[a,b]$ and differentiable on the open interval $(a,b)$, except possibly at the point $x_0$. Also assume that

$$ \lim_{x\to x_0^-}f'(x)=\lim_{x\to x_0^+}f'(x)=l. $$

We must prove that $f$ is differentiable at $x_0$ and that

\[ f'(x_0)=l. \]

First, choose a point $x<x_0$.

Since the function is continuous on the closed interval $[x,x_0]$ and differentiable on the open interval $(x,x_0)$, the Mean Value Theorem applies.

Therefore, there exists a point

\[ c\in(x,x_0) \]

such that

\[ \frac{f(x_0)-f(x)}{x_0-x}=f'(c). \]

Now let $x$ approach $x_0$ from the left.

The left-hand side becomes

\[ \lim_{x\to x_0^-}\frac{f(x_0)-f(x)}{x_0-x}=f'_-(x_0), \]

which is the left-hand derivative of $f$ at $x_0$.

Since $c$ always lies between $x$ and $x_0$, the point $c$ also approaches $x_0$ from the left as $x\to x_0^-$.

By assumption,

\[ \lim_{c\to x_0^-}f'(c)=l. \]

Therefore,

\[ f'_-(x_0)=l. \]

Applying the same argument to a point $x>x_0$ gives

\[ f'_+(x_0)=l. \]

The left-hand and right-hand derivatives both exist and are equal. Hence,

\[ f'(x_0)=l. \]

This completes the proof.

Geometric Interpretation. The derivative is the slope of the tangent line to the graph of a function. If the slopes of the tangent lines at points approaching $x_0$ converge to the same value from both the left and the right, then the tangent line at $x_0$ must have that same slope. Consequently, the function is differentiable at $x_0$.

And so on.