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I have two functions 1) $y=x^{x^{x}}$ 2) $y=(x^x)^x$

These two functions seem same to me and I just see it as a mere difference of writing style but when I graph it using an online graph plotter they have different curves also when I find their derivatives using logarithmic differentiation I get different results.For 1 and 2 I got $dy/dx$ as $x^{x^{x}}[x^x\cdot\ln(x)[1+\ln(x)]+x^{(x-1)}]$ and $(x^x)^x[x[2\ln(x)+1]]$ respectively

So,my question is ,Are these two functions really different,if yes ,how?If no,how can you justify their similar looking expressions?

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    They are indeed different. The second one is $x^{x^2}$. – 2017-02-13
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    @G.Sassatelli are x^x^x and (x^x)^x ,different for you? ,2nd function shown in question has Latex as (x^x)^x.Now what is your say? – 2017-02-13

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$x^{x^x}$ is normally parsed as $x^{(x^x)}$. Which is different than $(x^x)^x$ which is $x^{x^2}$.

To see the difference try $3^{3^3}=3^{27}$ whereas $(3^3)^3=27^3$. The first is on the order of $10^{12}$ where as the second is on the order of $10^5$.

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    I understand these facts sir but the second function has latex (x^x)^x.Still you will say it is equivalent to (x)^{x^2} ? – 2017-02-13
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    @KartikWatwani Yes. This is easiest seen on an example. Note that $(4^4)^4 = 4^4 * 4^4 * 4^4 * 4^4=4^{4+4+4+4}=4^{4*4}=4^{4^2}$ in this holds in general if you take $(n^n)^n$ that's the same $n^n\cdots n^n$ $n$ times which then is $n^{n+\cdots n}$ where the addition is $n$ times and so is $n^{n*n}=n^{n^2}$. This is obviously not even close to a proof but it gives you an inkling of the idea behind the definition. – 2017-02-13
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Note that the exponent of $y=x^{x^x}$ can be represented as $k=x^x$ so $y=x^k$. For $y=(x^x)^x$ the exponent inside of the bracket is multiplied with the outside exponent so $k=x\cdot x= x^2$ then $y=x^k=x^{x^2}$

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    I understand these facts sir but the second function has latex (x^x)^x.Still you will say it is equivalent to (x)^{x^2} ? – 2017-02-13
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    Both exponent are multiplied (exponent's rule) so $(x^x)^x = x^{x\cdot x} = x^{x^2}$. – 2017-02-13
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    but if I wanted to set (x^x) to the powered x .Then how will you express it? – 2017-02-13
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    $x^x$ to the $x-th$ power is $(x^x)^x$, as said before this yields $x^{x\cdot x}=x^{x^2}$. You could express $k=x^x$ then $k^x$ is what you are looking for. – 2017-02-13
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They are different, because $$3^{3^3} = 3^{(3^3)} = 3^{27} = 7\,625\,597\,484\,987$$ whilst $${(3^3)}^3 = 27^3 = 19\,683 = 3^9 = 3^{(3^2)}$$

One of general properties of exponentiation is $$(a^b)^c = a^{(b\cdot c)}$$ which corresponds to $$\log (a^b)^c = c\cdot\log a^b = c\cdot(b\cdot\log a) = (c\cdot b)\log a = \log a^{b\cdot c}$$ hence $$(x^x)^x = x^{(x\cdot x)} = x^{(x^2)} \ne x^{(x^x)}$$

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    I understand these facts sir but the second function has latex (x^x)^x.Still you will say it is equivalent to (x)^{x^2} ? – 2017-02-13
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    Sure it is. Please see my expanded answer. – 2017-02-13
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    but if I wanted to set (x^x) to the powered x .Then how will you express it? or does it mean x^(x^2) is same as saying (x^x) powered x? – 2017-02-13