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I have to prove that $\frac{d}{dx}a^x=a^x\cdot \log(a)$ for $a>0$ and $x>0$, using the difference quotient $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ but without using the chain rule.

I know that $\log(a^x)=x\cdot \log(a)$, and that for $y>0$ we have $y=\exp(\log(y))$, which means that $a^x=\exp(\log(a)\cdot x)$.

I have absolutely no idea how to solve this, as L'Hôpital's rule has yet to be covered and cannot be used here.

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P0lc3
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  • If you used LHopital’s rule, then you would take another derivative, so we definitely cannot use that. – Тyma Gaidash Dec 22 '21 at 23:26
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    Hint: $$ \frac{{a^{x + h} - a^x }}{h} = a^x \frac{{a^h - 1}}{h} = (a^x \log a) \times \frac{{e^{h\log a} - 1}}{{h\log a}}. $$ – Gary Dec 22 '21 at 23:28
  • Hint: $\lim_{h \to 0}\frac{e^h - 1}{h} = \frac{de^x}{dx}|_{x = 0}$ – Hermis14 Dec 22 '21 at 23:32
  • @Gary How do I solve $\frac{e^{h ;log(a)}-1}{h; log(a)}$?, that is not quite clear to me yet, because if that equals $1$ with $\lim_{h \to 0}$ that would get me to the assumption. – P0lc3 Dec 22 '21 at 23:41
  • Have you done implicit differentiation? Do you have to use the difference quotient? Because this is a pretty easy problem using implicit.

    (Hint: set $y=a^x$, then take $ln$ of both sides, then take $\frac{d}{dx}$ of both sides. You'll end up with a $\frac{1}{y}\cdot \frac{dy}{dx} = \dots$ and then plug in for $y$ and simplify a bit and you're done!)

     – mim Dec 22 '21 at 23:55
  • @mim Doesn't implicit differentiation make use of the chain rule? Yes, have to do it with difference quotient. – P0lc3 Dec 22 '21 at 23:59
  • $$ \mathop {\lim }\limits_{h \to 0} \frac{{e^{h\log a} - 1}}{{h\log a}} = \mathop {\lim }\limits_{h\log a \to 0} \frac{{e^{h\log a} - 1}}{{h\log a}} = \mathop {\lim }\limits_{h' \to 0} \frac{{e^{h'} - 1}}{{h'}} = 1, $$ see this. – Gary Dec 23 '21 at 00:26
  • @P0lc3 whoops! Good catch. I forgot that was a restriction. – mim Dec 23 '21 at 16:39

2 Answers2

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We have to show that $$\frac{d}{dx}a^x=a^x\log(a).$$

Proof: Recall that for any function $f(x)$, by definition, $$f'(x) =\lim _{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right), \forall f.$$ Let $f(x)=a^x$ s.t. $$\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}} \implies \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x+h}-a^{\displaystyle x}}{h}}. $$ $$\implies \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x} \times a^{\displaystyle h}-a^{\displaystyle x}}{h}}=\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \bigg(a^{\displaystyle x} \times \dfrac{a^{\displaystyle h}-1}{h}\bigg)}.$$ We can now evaluate $f'(x)$ by the constant multiple rule of calculus s.t. $$a^{\displaystyle x} \times\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle h}-1}{h}} \implies a^x\ln(a)$$ $$\therefore \forall f=a^x,f'(x)=a^x\ln(a)$$ Q.E.D

aygx
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  • I belive the problem of the OP is to show $(a^h-1)/h \to 1$. Your answer do not address this. – Gary Dec 23 '21 at 00:25
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you need to know that $$\lim_{x \rightarrow 0} \frac{e^x-1}x =1$$

which is proved here: proof of derivative of an exponential function

hellofriends
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