use logarithmic differentiation to find the derivative of the function y = (sqrt x)e^((x^2) - x) (x + 1)^(2/3)
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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))).`
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