Left-Hand Derivative

Consider a function defined on an interval extending to the left of the point x₀. We say that f(x) is left-differentiable at x₀ if the left-hand limit of the difference quotient between x and x₀ exists and is finite as x approaches x₀ from the left. $$ f'_-(x_0) = \lim_{x \rightarrow x_0^-} \frac{f(x)-f(x_0)}{x - x_0} $$ Since Δx = x − x₀, this can also be expressed as $$ f'_-(x_0) = \lim_{Δx \rightarrow 0^-} \frac{f(x_0 + Δx) - f(x_0)}{Δx} $$ which is known as the left-hand derivative of f(x).

Here, we’re focusing exclusively on values of x to the left of x₀.

When computing the left-hand derivative, the difference Δx = x − x₀ is always negative.

$$ x - x_0 < 0 $$

So,

$$ Δx < 0 $$

Note. Notation can vary depending on the author. Sometimes the point x is written as x₀ − δ. However, the idea is exactly the same: it refers to a point just to the left of x₀ that gets arbitrarily close to x₀ as we take the limit.

The interval to the right of x₀ isn’t considered here.

A Practical Example

Let’s check whether the left-hand derivative of the function f(x) = |x| exists at the point x₀ = 2.

$$ f(x) = 2 $$

We compute the limit as x approaches x₀ from the left:

$$ f'_-(x_0) = \lim_{x \rightarrow x_0^-} \frac{f(x) - f(x_0)}{x - x_0} $$

Substituting x₀ = 2, we have:

$$ f'_-(2) = \lim_{x \rightarrow 2^-} \frac{f(x) - f(2)}{x - 2} $$

$$ f'_-(2) = \lim_{x \rightarrow 2^-} \frac{|x| - |2|}{x - 2} $$

$$ f'_-(2) = \lim_{x \rightarrow 2^-} \frac{x - 2}{x - 2} = +1 $$

In this case, x is approaching 2 from the left without ever quite reaching it - for instance, values like 1.999999.

$$ f'_-(2) = \lim_{x \rightarrow 2^-} \frac{x - 2}{x - 2} = \lim_{x \rightarrow 2^-} \frac{1.99999 - 2}{1.99999 - 2} $$

So, the difference x − 2 is always negative in both the numerator and the denominator.

$$ f'_-(2) = \lim_{x \rightarrow 2^-} \frac{1.99999 - 2}{1.99999 - 2} = \lim_{x \rightarrow 2^-} \frac{-0.000001}{-0.000001} $$

Therefore, the quotient is always positive.

Since the values in the numerator and denominator are identical, the fraction simplifies to +1.

$$ f'_-(2) = \lim_{x \rightarrow 2^-} \frac{-0.000001}{-0.000001} = \lim_{x \rightarrow 2^-} 1 = +1 $$

In conclusion, the left-hand derivative of f(x) at x₀ = 2 is +1.

$$ f'_-(2) = +1 $$

Note. We can reach the same result using the alternative form of the left-hand limit: $$ f'_-(x_0) = \lim_{Δx \rightarrow 0^-} \frac{f(x_0 + Δx) - f(x_0)}{Δx} $$ $$ f'_-(2) = \lim_{Δx \rightarrow 0^-} \frac{f(2 + Δx) - f(2)}{Δx} $$ $$ f'_-(2) = \lim_{Δx \rightarrow 0^-} \frac{|2 + Δx| - |2|}{Δx} $$ For the left-hand limit, the increment Δx = x − x₀ is always negative, so Δx < 0. Thus, we can rewrite the expression as: $$ f'_-(2) = \lim_{Δx \rightarrow 0^-} \frac{|2 − Δx| − |2|}{− Δx} $$ The difference in the numerator, |2 − Δx| − |2|, is always negative. So, both the numerator and denominator are negative, which makes the overall quotient positive. $$ f'_-(2) = \lim_{Δx \rightarrow 0^-} \frac{- Δx}{- Δx} = \lim_{Δx \rightarrow 0^-} 1 = +1 $$ The result is the same either way.

And that’s how it works.