$4^{x^{x^x}}$
Hi, I came across this question and would like to check whether I have it done correctly: $e^{x^3}\ln4=4^{x^3}(3\ln4\cdot x^2)$ is this the correct solution?
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To kick start $$(f(x)^{g(x)})'=(\exp(g(x)\ln f(x))'=(\exp(g(x)\ln f(x))(g'(x)\ln f(x)+g(x)\cdot\frac1{f(x)}\cdot f'(x))$$
Vim
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Your order is wrong, as generally $a^{b^c}=a^{(b^c)}$(as noted by @AndréNicolas)
You should use that $(4^{f(x)})'=\ln 4 \times 4^{f(x)}f'(x)$.
Note that the derivative of $x^{g(x)}=x^{g(x)-1}(x \ln x g'(x) +g(x))$
As you pointed out, $(x^x)'=x^x(\ln x+1)$.
You can use these formulae to calculate $4^{x^{x^x}}$, which is $$4^{x^{x^x}}x^{x^x+x-1}\ln 4(x (\ln x)^2+x\ln x +1)$$
NOTE
You need not memorize the second formula, but it is not too difficult to calculate this, and you appear to be aware of how to do it.
S.C.B.
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$4^{x^{x^x}}$to show $4^{x^{x^x}}$. – Em. Feb 24 '16 at 00:26