Hi Kunle,

First you have to decide if this is

y = x^(xx)

or

y = (x^x)^x

Rather than making a choice I want to look at

y = x^f(x)

To differentiate this expression first take the natural logarithm of both sides to get

ln(y) = ln( x^f(x) ) = f(x) ln(x)

Differentiating both sides gives

 ^y'/_y = f'(x) ln(x) + f(x) ¹/_x

Thus

y' = y ( f'(x) ln(x) + f(x) ¹/_x) = x^f(x) ( f'(x) ln(x) + f(x) ¹/_x )

This should allow you to answer your queation.

Penny

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