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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, 2015 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, 2015 at 13:58
  • $\begingroup$ explain first principle??? $\endgroup$ May 15, 2015 at 13:59
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    $\begingroup$ $$\lim_{h\to0}\frac{{f(x+h)}}-f(x)}{h}$$ $\endgroup$ May 15, 2015 at 14:03
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    $\begingroup$ yeah, like f(x+h) - f(x) /h as h -> 0 $\endgroup$ May 15, 2015 at 14:04

5 Answers 5

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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 ${7^{ln(x)}*lnx}/x$ $\endgroup$ May 15, 2015 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, 2015 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, 2015 at 5:55
  • $\begingroup$ I doubt that generalized binomial series counts as "first principles". $\endgroup$
    – Wojowu
    May 16, 2015 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, $\endgroup$ May 15, 2015 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, 2015 at 14:06
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    $\begingroup$ Yeah, Well I want the steps to solve that limit I gave basicallY $\endgroup$ May 15, 2015 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, 2015 at 14:07
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    $\begingroup$ @user46944 thanks. Corrected. $\endgroup$
    – Umberto P.
    May 15, 2015 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, $\endgroup$ May 15, 2015 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, 2015 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. $\endgroup$ May 18, 2015 at 17:45

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