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If y =x^(x+1),then y'=?
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You need to use logarithmic differentiation such that:
`ln y = ln x^(x+1) => ln y = (x + 1)*ln x`
You need to differentiate both sides, such that:
`(1/y)*y' = (x + 1)'*ln x + (x + 1)*(ln x)'`
`y' = y*(ln x + (x+1)/x)`
Replacing `x^(x+1)` for y yields:
`y' = x^(x+1)*(ln x + 1 + 1/x)`
Hence, evaluating the derivative y', using logarithmic differentiation, yields `y' = x^(x+1)*(ln x + 1 + 1/x).`
Posted by sciencesolve on July 4, 2013 at 5:53 PM (Answer #1)
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