Question

lf y = `2^(x^(2^(x^(...∞))))`, then x(1 - y logx logy)`dy/dx` = ______  

Options

• y² logy 

• y log y

• `y^2/logy`

• `y/logy`

Answer 1

lf y = `2^(x^(2^(x^(...∞))))`, then x(1 - y logx logy)`dy/dx` = y² logy

Explanation:

y = `2^{x^y}`

∴ log y = x^y log 2

∴ log(log y) = y logx + log(log 2)

Differentiating both sides w.r.t.x, we get

`1/logy . 1/y . dy/dx = y/x + dy/dx logx`

⇒ `(1/(ylogy) - logx)dy/dx = y/x`

⇒ x(1 - y logx logy) `dy/dx = y^2 logy`