# If y =x^(x+1),then y'=?

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### 1 Answer

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).`

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