Hi, this was on my last calc test, and I wanted to make sure I understood why I got it wrong.
Find the derivative of
(2^{2}^{^x})
My answer:
ln(2)2^{2^x}
My guess now that I realize it's wrong:
ln(4)(4^{x})
Thanks!
Hello, rocapp!
This requires some fancy footwork . . .
$\displaystyle \text{Find the derivative of: }\:y \;=\;2^{2^x}$
$\displaystyle \text{Take logs: }\:\ln(y) \:=\:\ln\left(2^{2^x}\right) \quad\Rightarrow\quad \ln{y}\;=\;2^x\!\cdot\!\ln 2$
$\displaystyle \text{Differentiate implicitly: }\:\frac{1}{y}\!\cdot\!y' \;=\;2^x\!\cdot\!(\ln 2)^2 $
$\displaystyle \text{Therefore: }\:y' \;=\;y\!\cdot\!2^x\!\cdot\!(\ln2)^2 \quad\Rightarrow\quad y'\;=\;2^{2^x}\!\cdot2^x\!\cdot\!(\ln2)^2$