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Question
If y = `x^(x^(x^(.^(.^.∞))`, then show that `"dy"/"dx" = y^2/(x(1 - logy).`.
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Solution
y = `x^(x^(x^(.^(.^.∞))`
∴ log y = `log(x^(x^(x^(.^(.^.∞)))))`
= `x^(x^(x^(.^(.^.∞)))).logx`
∴ log y = y log x ...(1)
Differentiating both sides w.r.t. x, we get
`(1)/y.dy/dx = y.d/dx(logx) + (logx)dy/dx`
∴ `(1)/ydy/dx = y xx (1)/x + (logx)dy/dx`
∴ `(1/y - logx)dy/dx = y/x`
∴ `((1 - ylogx)/(y))dy/dx"= y/x`
∴ `dy/dx = y^2/(x(1 - ylogx)`
∴ `dy/dx = y^2/(x(1 - logy)`. ...[By (1)]
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