You need to take logarithms both sides such that:

`ln y = ln((sqrt x)e^((x^2) - x)(x + 1)^(2/3))`

Using logarithmic identities yields:

`ln y = ln sqrt x + ln e^(x^2 - x) + ln (x+1)^(2/3)`

`ln y = (1/2)ln x + (x^2 - x) ln e + (2/3) ln...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

You need to take logarithms both sides such that:

`ln y = ln((sqrt x)e^((x^2) - x)(x + 1)^(2/3))`

Using logarithmic identities yields:

`ln y = ln sqrt x + ln e^(x^2 - x) + ln (x+1)^(2/3)`

`ln y = (1/2)ln x + (x^2 - x) ln e + (2/3) ln (x+1)`

Since `ln e = 1` yields:

`ln y = (1/2)ln x + (x^2 - x) + (2/3) ln (x+1)`

Differentiating both sides with respect to x yields:

`(1/y)*y' = 1/(2x) + 2x - 1 + 2/(3(x+1))`

`y' = y(1/(2x) + 2x - 1 + 2/(3(x+1))) `

Substituting `((sqrt x)e^((x^2) - x)(x + 1)^(2/3))` for y yields:

`y' = ((sqrt x)e^((x^2) - x)(x + 1)^(2/3))*(1/(2x) + 2x - 1 + 2/(3(x+1))) `

**Hence, diferentiating the given function using logarithmic differentiation yields `y' = ((sqrt x)e^((x^2) - x)(x + 1)^(2/3))*(1/(2x) + 2x - 1 + 2/(3(x+1))).` **