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I still can't figure this out,

Question is Find the Derivative $ 7^{\ln(x)} $ using first principle

This is where I got

$$\lim_{h \to 0} \frac{7^{\ln(x+h)} - 7^{\ln(x)}}{h} $$

then What should I do?

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    $\begingroup$ Mind explaining the first principle? $\endgroup$ – user228113 May 15 '15 at 13:57
  • $\begingroup$ @CuriousSciDude We will be better able to help you with the question if you explain what the first principle is. Please update your question with this info ASAP so you have the best chance of being helped. $\endgroup$ – layman May 15 '15 at 13:58
  • $\begingroup$ explain first principle??? $\endgroup$ – CuriousSciDude May 15 '15 at 13:59
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    $\begingroup$ $$\lim_{h\to0}\frac{{f(x+h)}}-f(x)}{h}$$ $\endgroup$ – CuriousSciDude May 15 '15 at 14:03
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    $\begingroup$ yeah, like f(x+h) - f(x) /h as h -> 0 $\endgroup$ – CuriousSciDude May 15 '15 at 14:04
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Note that calculating derivative of $f(x)$ via first principles means that we need to calculate the limit $$\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$ without using any rules of differentiation.

Here $f(x) = 7^{\log x}$ and we can proceed as follows \begin{align} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\ &= \lim_{h \to 0}\frac{7^{\log(x + h)} - 7^{\log x}}{h}\notag\\ &= 7^{\log x}\lim_{h \to 0}\frac{7^{\log(x + h) - \log x} - 1}{h}\notag\\ &= 7^{\log x}\lim_{h \to 0}\frac{7^{\log(x + h) - \log x} - 1}{\log(x + h) - \log x}\cdot\frac{\log(x + h) - \log x}{h}\notag\\ &= 7^{\log x}\lim_{h \to 0}\frac{7^{\log(x + h) - \log x} - 1}{\log(x + h) - \log x}\cdot\lim_{h \to 0}\frac{\log(x + h) - \log x}{h}\notag\\ &= 7^{\log x}\lim_{t \to 0}\frac{7^{t} - 1}{t}\cdot\lim_{h \to 0}\frac{\log((x + h)/x)}{h}\text{ (by putting }t = \log(x + h) - \log x)\notag\\ &= 7^{\log x}\lim_{t \to 0}\frac{e^{t\log 7} - 1}{t}\cdot\lim_{h \to 0}\frac{\log(1 + (h/x))}{h}\notag\\ &= 7^{\log x}\lim_{t \to 0}\log 7 \cdot\frac{e^{t\log 7} - 1}{t\log 7}\cdot\lim_{h \to 0}\frac{\log(1 + (h/x))}{h/x}\cdot\frac{1}{x}\notag\\ &= \frac{7^{\log x}\log 7}{x}\lim_{y \to 0}\cdot\frac{e^{y} - 1}{y}\cdot\lim_{z \to 0}\frac{\log(1 + z)}{z}\text{ (putting }y = t\log 7, z = h/x)\notag\\ &= \frac{7^{\log x}\log 7}{x}\cdot 1\cdot 1\notag\\ &= \frac{7^{\log x}\log 7}{x} \end{align}

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Note that $7^{\ln(x)} = \left(e^{\ln(7)} \right)^{\ln(x)} = \left(e^{\ln(x)} \right)^{\ln(7)} = x^{\ln(7)}$.

From the generalized binomial theorem, we have $$(x+h)^{\alpha} = \sum_{k=0}^{\infty} \dbinom{\alpha}k x^{\alpha-k}h^k = x^{\alpha} + \alpha x^{\alpha-1}h + h^2 f(x,h;\alpha)$$ where $f(x,0;\alpha)$ is continuous in $h$ with $f(x,0;\alpha) = 0$. Hence, $$\dfrac{(x+h)^{\alpha}-x^{\alpha}}h = \alpha x^{\alpha-1} + hf(x,h;\alpha)$$ Hence, we have $$\lim_{h \to 0}\dfrac{(x+h)^{\alpha}-x^{\alpha}}h = \alpha x^{\alpha-1} + \lim_{h \to 0}hf(x,h;\alpha) = \alpha x^{\alpha-1}$$

Hence, we have $$\lim_{h \to 0} \dfrac{7^{\ln(x+h)}-7^{\ln(x)}}h = \lim_{h \to 0} \dfrac{(x+h)^{\ln(7)}-x^{\ln(7)}}h = \ln(7)x^{\left(\ln(7)-1\right)}$$

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  • $\begingroup$ But the answer is ${7^{ln(x)}*lnx}/x$ $\endgroup$ – CuriousSciDude May 15 '15 at 14:17
  • $\begingroup$ @CuriousSciDude: Answer given here is correct and if you like it your way it should be visible as $7^{\log x}\log 7/x$ and not $7^{\log x}\log x/x$. $\endgroup$ – Paramanand Singh May 16 '15 at 5:53
  • $\begingroup$ I like your use of $a^{\log b} = b^{\log a}$ equality, but since my daily voting limit is reached I can't upvote your answer. will do that tomorrow. $\endgroup$ – Paramanand Singh May 16 '15 at 5:55
  • $\begingroup$ I doubt that generalized binomial series counts as "first principles". $\endgroup$ – Wojowu May 16 '15 at 6:35
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I'm not sure if this is "first principle" or not, but since $7^a = e^{(\ln x)(\ln 7)}$ for any real $a$ you have $$7^{\ln x} = e^{(\ln x)(\ln 7)}.$$ Thus $$\lim_{h \to 0} \frac{7^{\ln(x+h)} - 7^{\ln x}}{h} = \frac d{dx} e^{(\ln x)(\ln 7)}$$ which you can compute using the usual rules of differentiation.

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  • $\begingroup$ I might try this one lol, $\endgroup$ – CuriousSciDude May 15 '15 at 14:05
  • $\begingroup$ I think using "first principle" would mean taking the limit and show that it matches the "usual rules of differentiation". $\endgroup$ – CSCFCEM May 15 '15 at 14:06
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    $\begingroup$ Yeah, Well I want the steps to solve that limit I gave basicallY $\endgroup$ – CuriousSciDude May 15 '15 at 14:07
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    $\begingroup$ I don't think this is quite right: $7^{\ln(x)}=e^{\ln(7) \ln(x)}$. $\endgroup$ – Ian May 15 '15 at 14:07
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    $\begingroup$ @user46944 thanks. Corrected. $\endgroup$ – Umberto P. May 15 '15 at 14:09
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This is the derivative of $(7)^{\ln x}$ or the derivative of $e^{\ln x\cdot \ln 7}$ or the derivative of $x^{\ln 7}$ which is $\ln 7 \cdot x^{(\ln 7) -1}=(1/x)\ln7\cdot x^{\ln7}$.

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  • $\begingroup$ I wanna solve that limit dude, $\endgroup$ – CuriousSciDude May 15 '15 at 14:11
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$$\lim_{h \to 0} \frac{7^{\ln(x+h)}-7^{\ln(x)}}{h}=$$ $$7^{\ln(x)}\lim_{h \to 0} \frac{7^{\ln(x+h)-\ln(x)}-7^{\ln(x)-\ln(x)}}{h}$$ $$7^{\ln(x)}\lim_{h \to 0} \frac{7^{\ln(1+h/x)}-1}{h}$$ I use the Maclaurin series expansion of $\ln(1+x)=x-x^2/2+x^3/3-x^4/4+\,\dots$: $$7^{\ln(x)}\lim_{h \to 0} \frac{7^{h/x}-1}{h}$$ $$7^{\ln(x)}\lim_{h \to 0} \frac{(7^{1/x})^h-1}{h}$$ Now I use the well-known limit $\displaystyle\lim_{h \to 0}\frac{a^h-1}{h}=\ln(a)$:

$$7^{\ln(x)}\ln(7^{1/x})=\frac{7^{\ln(x)}\ln(7)}{x}$$

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  • $\begingroup$ you have a curious fallacy where you replace $\log(1 + (h/x))$ with $h/x$. There is no general rule which allows such replacements. $\endgroup$ – Paramanand Singh May 16 '15 at 10:00
  • $\begingroup$ @Paramanand: This is indeed true. It had to use the Maclaurin series approximation, otherwise I was stuck. Fortunately, I could get the correct derivative. I admit your reasoning is far better. $\endgroup$ – Steven Van Geluwe May 18 '15 at 17:45

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