I encountered this problem in high school math textbook.

It was asking to show that if $f(x) = a^{x}$ then show that $f'(x)=a^{x}\ln(a)$ using the definition of the derivative.

I'm only aware of how to prove this relationship by using taylor series or the identity

$$e^{x}=\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n$$

I was wondering can someone come up with a way simple to prove it so that a high school student without understanding of taylor series or exponential identities can understand the proof?

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    $\begingroup$ "without ... exponential identities..." that's a tall order. If this high school student in question understands the basics of derivatives, I imagine exponential identities are also within reach $\endgroup$ – jameselmore Apr 29 '17 at 19:07
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    $\begingroup$ Well, first you have to decide how to define $a^x$ and $\ln(x)$. $\endgroup$ – carmichael561 Apr 29 '17 at 19:07
  • $\begingroup$ Your limit is irrelevant, because you want to calculate $\frac{|a^t -a^x|}{|t-x|} $ as t getting close to x. $\endgroup$ – Yujie Zha Apr 29 '17 at 19:08
  • $\begingroup$ @carmichael561 Its from a high school math textbook. I'm not sure if the textbook has defined $a^{x}$ or $ln(x)$. $\endgroup$ – user262291 Apr 29 '17 at 19:12
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    $\begingroup$ It is a common sin committed by many textbook authors who do not define $a^{x} $ and $\log x$ and yet expect the readers to have a sound understanding of the limit in question here. Moreover defining such things is hard and it is better to present these things without proof to a high school student. After all no one raises an eyebrow when a 12 year old gets to know that $\pi$ is irrational (without proof). $\endgroup$ – Paramanand Singh Apr 29 '17 at 20:16

we denote $$f(x)=a^x$$ taking the logarithm on both sides we obtain: $$\ln(f(x))=x\ln(a)$$ and by the cain rule we get $$\frac{f'(x)}{f(x)}=\ln(a)$$ and multiplying by $$f(x)$$ we get the desired result.

or $$\frac{a^{x+h}-a^x}{h}=a^x\left(\frac{a^h-1}{h}\right)$$ and use that the limit in brackets is equal to $\ln(a)$ for $a>0$

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    $\begingroup$ i have it provided $\endgroup$ – Dr. Sonnhard Graubner Apr 29 '17 at 19:13
  • $\begingroup$ Now the question that can you prove $\lim_{h\to 0}{e^h - 1\over h } = 1$ without taylor series ? $\endgroup$ – A---B Apr 29 '17 at 19:16
  • $\begingroup$ to prove this limit set $$t=a^h-1$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 29 '17 at 19:17


Note that $$\lim_{h\to0}\frac{e^h-1}{h}=1$$


$$\lim_{h\to0}\frac{a^h-1}{h}=\ln a\left(\lim_{h\to0}\frac{e^{h\ln a}-1}{h\ln a}\right)=\ln a$$


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