Question

LGBADERIVATIVE:

\[\huge y = 2^{3^{x^2}}\]

Answer 1

im thinking ln?

Answer 2

\[\ln y = 3^{x^2} \ln 2\]

Answer 3

\[\huge \frac 1y y' = 2x^2 3^{x^2} \ln 2\]
is that right?

Answer 4

\[\huge y' = (2^{3^{x^2}} )(2x^2)(3^{x^2})(\ln 2)\]

Answer 5

did i do that right...

\[a = 3^{x^2}\]
\[\ln a = x^2 \ln 3\]
\[\frac 1a a' = 2x \ln 3\]
\[a' = 3^{x^2} 2x\ln 3\]
oops

Answer 6

dy/dx= 2^(3x^2) * ln2 * ln 3 *2x? o.0

Answer 7

my answer is

\[\huge y' = (2^{3^{x^2}})(3^{x^2} 2x\ln 3)(\ln 2)\]
that right?

Answer 8

thats what i got @lgbasallote :o

Answer 9

haha yay...wonder if we're correct

Answer 10

me too
\[y^\prime=(2^{3^{x^2}})(\ln{2})(3^{x^2})(\ln{3})2x\]

Answer 11

ahh then we're all right since it's unanimous!!

Answer 12

You are going right @lgbasallote