Why can't derivatives jump over 0?

Question

Derivatives follow $nx^{n-1}$ which works for either positive numbers or negative numbers. For example:

With $f_1(x)=X^3$ then $f_1'(x)=3X^2$

With $f_2(x)=X^{-3}$ then $f_2'(x)=-3X^{-4}$

But other than the obvious (0 times anything is 0), why can't derivatives jump over 0?

Answer 1

For any $\alpha\in\mathbb{R}\backslash\{0\}$ and $f(x)=x^\alpha$ you find

$$f'(x)=\alpha x^{\alpha-1}$$

Especially for any $\alpha\in(0,1)$, you find a "jump over $0$" because $\alpha-1 < 0$.

---

Note that for $\alpha=0$, it holds $f(x)=x^\alpha =1$ with

$$f'(x)=0$$

In this case, the derivative is not a non-trivial multiple of $x^{\alpha-1}=x^{-1}$.

However, there exists an antiderivative for $x^{-1}$: For $f(x)=\ln(|x|)$, it holds

$$f'(x)=x^{-1}$$

Answer 2

Restatement of the Question to Make it Easier to Reference in the Answer

Derivatives follow $nx^{n-1}$ which works for either positive numbers or negative numbers. For example:

With $f_1(x)=X^3$ then $f_1'(x)=3X^2$

With $f_2(x)=X^{-3}$ then $f_2'(x)=-3X^{-4}$

But other than the obvious (0 times anything is 0), why can't derivatives jump over 0?

Solution Steps

The proof for:

$$\boxed{ \left(x^n\right)^{'}=\lim_{h \to 0}\frac{(x+h)^n-x^n}{h}=n \, x^{n-1} }$$

is in my answer to the Math Stack Exchange article "Proof in the Derivative of $x^n\,\,$"

When $x$ is $x\overset{\text{is the constant}}=0$, then $\left(x^n\right)^{'}=0$ for $n\ge 0$. Smooth continuous functions can be built up as a Taylor Series (from Taylor's Theorem) as $f=\sum_{i\ge 0} a_i x^i$, where it is "an approximation of a $ {\textstyle k}-$ times differentiable function around a given point by a polynomial of degree ${\textstyle k}$, called the ${\textstyle k}$-th-order Taylor polynomial"

So in a sense as long as there is no discontinuous point at $x=0$, then "jumping over" is just taking a limit of the function at $f\left(x\right) \to f\left(x_0\right)$ and $f\left(x+h\right) \to f\left(x_0+h\right)$.

The problem is when $f\left(0\right)$ is not defined, or when it is not continuous, resulting in a finite (or unbounded infinite) value in the top of the fraction namely $f\left(x=0+h\right)-f\left(x=0\right)$. In that case, there is no (finite) limit as $\frac1{h}\to \frac1{h\to 0} \to \infty$.

Answer 3

Let's return to your examples:

• First, you consider $f_1(x) = x^3$. This function is continuous at 0. It's derivative is $3x^2$ which is also continuous at 0. This function has an ordinary derivative at 0.

• Second, you consider $f_2(x) = x^{-3}$. This function is not continuous at 0:

  • coming in from the right, that is: $x \approx 0$ and $x > 0$, it is tending to $\infty$.
  • coming in from the left, that is $x \approx 0$ and $x < 0$, it is tending to $-\infty$.

  Since the function does not have a value at 0, it does not have an ordinary derivative at 0. I won't try to make a mathematically precise version of that statement since I do not know your level, but I encourage you to return to whichever definition of differentiability you have and see that it does not apply to $f_2$ at $x=0$