Differentiating $4^{x^{x^x}}$

Question

$4^{x^{x^x}}$

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?

Answer 1

To kick start $$(f(x)^{g(x)})'=(\exp(g(x)\ln f(x))'=(\exp(g(x)\ln f(x))(g'(x)\ln f(x)+g(x)\cdot\frac1{f(x)}\cdot f'(x))$$

Answer 2

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.