# How do you find the derivative of 2^(3^(x^2))?

Jun 30, 2015

Well, first let's make it look nicer...

$y = {2}^{{3}^{{x}^{2}}}$

Let $\textcolor{\mathrm{da} r k red}{u = {3}^{{x}^{2}}}$

$\implies y = {2}^{u}$

Using logarithmic simplification:
${\ln}_{2} \left(y\right) = u \to \frac{\ln y}{\ln 2} = u \to \ln y = u \ln 2$

And now with some manipulation:
$y = {e}^{u \ln 2} = {e}^{\ln {2}^{u}} = {2}^{u}$

$\textcolor{g r e e n}{\frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u \ln 2} \cdot \ln 2 = {2}^{u} \ln 2}$

Similarly:

Let $\textcolor{\mathrm{da} r k red}{w = {x}^{2}}$

$\implies u = {3}^{w}$

${\ln}_{3} \left(u\right) = w \to \frac{\ln u}{\ln 3} = w \to \ln u = w \ln 3$

$u = {e}^{w \ln 3} = {e}^{\ln {3}^{w}} = {3}^{w}$

$\textcolor{g r e e n}{\frac{\mathrm{du}}{\mathrm{dw}} = {e}^{w \ln 3} \cdot \ln 3 = {3}^{w} \ln 3}$

Lastly, $\textcolor{g r e e n}{\frac{\mathrm{dw}}{\mathrm{dx}} = 2 x}$.

Thus, taking the whole derivative:

$\frac{d}{\mathrm{dx}} \left[y = {2}^{u \left(w \left(x\right)\right)}\right] = \frac{d}{\mathrm{dx}} \left[y = {2}^{u \left({x}^{2}\right)}\right] = \frac{d}{\mathrm{dx}} \left[y = {2}^{{3}^{{x}^{2}}}\right] = \left(\frac{\mathrm{dy}}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dw}}\right) \left(\frac{\mathrm{dw}}{\mathrm{dx}}\right)$

$= \left({2}^{u} \ln 2\right) \left({3}^{w} \ln 3\right) \left(2 x\right)$

$= \left({2}^{{3}^{{x}^{2}}} \ln 2\right) \left({3}^{{x}^{2}} \ln 3\right) \left(2 x\right)$

$= 2 x \ln 2 \ln 3 \left({2}^{{3}^{{x}^{2}}} {3}^{{x}^{2}}\right)$

$= \textcolor{b l u e}{x \ln 2 \ln 3 \left({2}^{{3}^{{x}^{2}} + 1} {3}^{{x}^{2}}\right)}$