How do you differentiate #y=2^(3^(x^2))#?

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Answer 1 · 2017-01-30T16:26:43

#(dy)/(dx)=2^(3^(x^2))xx3^(x^2)ln18xx x#

As #y=2^(3^(x^2)#, we have

#lny=3^(x^2)ln2# or #3^(x^2)=lny/ln2#

i.e. #x^2ln3=ln(lny/ln2)#

and hence differentiating we get

#2ln3xx x=1/((lny/ln2))xx1/ln2xx1/y(dy)/(dx)#

or #2ln3xx x=1/lnyxx1/y(dy)/(dx)#

and #(dy)/(dx)=yxxlnyxx2ln3xx x#

= #2^(3^(x^2))xx3^(x^2)ln2xx2ln3xx x#

= #2^(3^(x^2))xx3^(x^2)ln18xx x#