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Could anyone help with this question:

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Edit: Reducing it to one of these options would be great!

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closed as off-topic by user91500, zoli, R_D, user1551, mrs Apr 28 '17 at 12:07

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Think of $y$ as being defined by the implicit equation $\,y = a^{x^y}\,$. $\endgroup$ – dxiv Apr 28 '17 at 4:02
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    $\begingroup$ Disregarding any questions of convergence, you would have $\log(\log(y)) = y\log(x) + \log(\log(a))$. Then using implicit differentiation we have $\frac{y'}{y\log(y)} = y' \log(x) + \log(\log(a))$. Then $y' = \frac{y\log(y)\log(\log(a))}{1 - y\log(y)\log(x)}$ $\endgroup$ – Tob Ernack Apr 28 '17 at 4:02
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    $\begingroup$ Is it $a$ raised to a chain of $x$ copies, or $a^{x}$ raised by a chain of $a^{x}$ copies? $\endgroup$ – dantopa Apr 28 '17 at 4:02
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    $\begingroup$ Actually I made a mistake in my implicit differentiation... You get $\frac{y'}{y\log(y)} = y'\log(x) + \frac{y}{x}$ and so $y' = \frac{y^2\log(y)}{x(1 - y\log(y)\log(x))}$ $\endgroup$ – Tob Ernack Apr 28 '17 at 4:15
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    $\begingroup$ @MichaelHoppe The posted expression is usually parsed as $a^{(x^{(y)})}=a^{x^y}\,$. $\endgroup$ – dxiv Apr 28 '17 at 6:58
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I will assume that everything is well defined and real, not paying attention to whether logarithms are taken of positive numbers and such.

We have $y=a^{x^y}$ (not $y=(a^x)^y$ which would mean $y=a^{{xa}^{{xa}^{\cdots}}}$). Let me denote the derivative operator by $D$.

First we need a little lemma: If $f$ and $g$ are functions of $x$, then $$ D(f^g) = De^{g\log(f)} = e^{g\log(f)}D(g\log(f)) = f^g(g'\log(f)+gf'/f). $$ (There is an easy way to memorize this. If $f$ is constant, the derivative is $f^g\log(f)g'$. If $g$ is constant, the derivative is $gf^{g-1}$. The full derivative is the sum of these two. This argument can be formalized, but it'd be a sidetrack here.)

Applying this to $f(x)=x$ and $g(x)=y(x)$ gives $$ D(x^y) = x^y(y'\log(x)+y/x). $$ This will be useful soon.

Taking the derivative gives \begin{align} y' &= D(a^{x^y}) \\&= a^{x^y}\log(a)D(x^y) \\&= y\log(a)D(x^y) %\\&= %y\log(a)[yx^{y-1}+\log(x)x^yy'] \\&= y\log(a)x^y[y/x+\log(x)y'] \\&= y\log(a^{x^y})[y/x+\log(x)y'] \\&= y\log(y)[y/x+\log(x)y'] . \end{align} This gives $$ [1-\log(x)y\log(y)]y' = y^2\log(y)/x, $$ from which we can solve $$ y'=\frac{y^2\log(y)}{x[1-\log(x)y\log(y)]}. $$ This does not seem to match any of the options given. Here is an alternative form, using $x^y=\log(y)/\log(a)$: $$ y'=\frac{y^2\log(y)}{x\left[1-\log\left(\frac{\log(y)}{\log(a)}\right)\log(y)\right]}. $$

Perhaps I miss a way to manipulate the formula, or perhaps the problem is mistaken; as others have pointed out, taking $y=(a^x)^y$ leads to option C.

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  • $\begingroup$ Thanks for this lemma: $D(f^g)=De^{g\log(f)}=e^{g\log(f)}D(g\log(f))=f^g(g'\log(f)+gf'/f)$; I've done many derivatives but didn't come across this simple little trick, really appreciate this :) $\endgroup$ – k.Vijay Apr 28 '17 at 8:50
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    $\begingroup$ @k.Vijay You are welcome! :) That's actually an example of a more general method of differentiation, which makes differentiation of almost any function easy. This general thing is essentially my parenthetical remark after the lemma: you can treat different functions or instances of $x$ as constants in a certain way. $\endgroup$ – Joonas Ilmavirta Apr 28 '17 at 8:57
  • $\begingroup$ yes, this is more generalization which helps to solve problem like this (when both are variable), btw using this generalization we can do any kind of differentiation of exponential function... $\endgroup$ – k.Vijay Apr 28 '17 at 9:20
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\begin{align*} y&=a^{x^{a^{x\cdots}}}\\ \implies y&=a^{x^y}\\ \implies \ln y&=x^y\ln a\hspace{25pt}\cdots\text{(i)}\\ \implies \dfrac{1}{y}\dfrac{dy}{dx}&=\ln a\cdot\dfrac{d}{dx}\left(x^y\right)\\ &=\ln a\cdot\dfrac{d}{dx}\left(e^{y\ln x}\right)\\ &=\ln a\cdot e^{y\ln x}\cdot\dfrac{d}{dx}(y\ln x)\\ &=\ln a\cdot x^y\cdot\left(\dfrac{y}{x}+\ln x\cdot\dfrac{dy}{dx}\right)\\ \implies \dfrac{dy}{dx}&=y\cdot x^y\ln a\cdot\left(\dfrac{y}{x}+\ln x\cdot\dfrac{dy}{dx}\right)\\ \implies \dfrac{dy}{dx}\left(1-y\cdot x^y\ln a\cdot\ln x\right)&=\dfrac{y^2}{x}\cdot x^y\ln a\\ \implies \dfrac{dy}{dx}\left(1-y\cdot\ln y\cdot\ln x\right)&=\dfrac{y^2}{x}\cdot\ln y\hspace{25pt}\text{ as from (i) }[x^y\ln a=\ln y]\\ \implies \dfrac{dy}{dx}&=\dfrac{y^2\ln y}{x\left(1-y\cdot\ln y\cdot\ln x\right)} \end{align*} So, this doesn't matches any answer.

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    $\begingroup$ You've made a mistake in the second line; I've corrected that. $\endgroup$ – Michael Hoppe Apr 28 '17 at 5:26
  • $\begingroup$ ohh .. yes yes .. thanks for that :) $\endgroup$ – k.Vijay Apr 28 '17 at 5:33
  • $\begingroup$ You're welcome. $\endgroup$ – Michael Hoppe Apr 28 '17 at 6:17
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    $\begingroup$ Should not it be $\displaystyle a^{x^{y}}$ in the second line not $a^{xy}$ ? $\endgroup$ – A---B Apr 28 '17 at 7:33
  • $\begingroup$ ohoo..!!! I've made a huge mistake, thanks for pointing this out @A---B, let me try to fix this $\endgroup$ – k.Vijay Apr 28 '17 at 8:39
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Building blocks: $$ D\left( a^{t(x)}, x\right) = a^x \ln (a) t'(x), \qquad D\left( x^x, x\right) = x^x (1 + \ln(x)) $$

First chains $$ \begin{align} D\left( a^{x}, x\right) &= a^x \ln (a) \\ D\left( a^{x^{x}}, x\right) &= x^x a^{x^x} \ln (a) (\ln (x)+1) \\ D\left( a^{x^{x^{x}}}, x\right)&= a^{x^{x^x}} x^{x^x} \ln (a) \left(x^{x-1}+x^x \ln (x) (\ln (x)+1)\right) \\ D\left( \left( \left( \left( a^x \right)^x \right)^x \right)^x, x \right) &= \left(\left(\left(a^x\right)^x\right)^x\right)^x \left(x \left(x \left(\ln \left(a^x\right)+x \ln (a)\right)+\ln \left(\left(a^x\right)^x\right)\right)+\ln \left(\left(\left(a^x\right)^x\right)^x\right)\right) \end{align} $$

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    $\begingroup$ OP's question is not about $a^{x^{\color{red}{x}^\cdots}}$. $\endgroup$ – dxiv Apr 28 '17 at 7:45

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