关于用公式$d(a^x)/dx = a^x \ln x$求$y=1^x$导数的疑问及替代方法咨询

Hey there! First off, let's spot a tiny typo in the formula you mentioned—it should be $d(a^x)/dx = a^x \ln a$, not $\ln x$. That's probably just a slip-up, but it's good to clarify first.

Now, why might you hesitate to use this formula for $y=1^x$? Let's break it down:

• The standard exponential derivative rule $d(a^x)/dx = a^x \ln a$ is designed for positive real numbers $a$ where $a \neq 1$. When $a=1$, the function $1^x$ isn't a "true" exponential function—it's just a constant function equal to 1 for every real $x$.
• If you do plug $a=1$ into the formula, you get $1^x \cdot \ln 1 = 1 \cdot 0 = 0$, which is actually the correct derivative. But relying on the formula here misses the simpler, more fundamental point: this is a constant, so its derivative is zero by definition.

As for other ways to differentiate $y=1^x$, here are a few straightforward methods:

• Simplify first (the easiest approach): For any real value of $x$, $1^x = 1$. The derivative of a constant function is always 0, so we can immediately say $y' = 0$.
• Logarithmic differentiation (overkill but valid): Take the natural log of both sides: $\ln y = \ln(1^x) = x \cdot \ln 1 = 0$. Differentiate both sides with respect to $x$: $\frac{y'}{y} = 0$. Multiply both sides by $y$ (which is 1) and you get $y' = 0$.
• Limit definition (going back to basics): Use the formal definition of the derivative:
  $$
  y' = \lim_{h \to 0} \frac{1^{x+h} - 1^x}{h}
  $$
  Since $1^{x+h} = 1$ and $1^x = 1$, this simplifies to:
  $$
  \lim_{h \to 0} \frac{1 - 1}{h} = \lim_{h \to 0} \frac{0}{h} = 0
  $$

备注：内容来源于stack exchange，提问作者Shinnaaan