Question

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

Answer 1

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

Explanation:

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`