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

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### 2 Answers

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

Taking the natural logarithm of both the sides and applying the properties of the logarithms, we get

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

Differentiating both sides with respect to x

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

`1/y dy/dx=(3(x+1)+6x(x+1)(2x-1)+4x)/(6x(x+1))`

`1/ydy/dx=(3x+3+12x^3-6x^2+12x^2-6x+4x)/(6x(x+1))`

`1/ydy/dx=(12x^3+6x^2+x+3)/(6x(x+1))`

`dy/dx=(sqrt(x)e^(x^2-x) (x+1)^(2/3))(12x^3+6x^2+x+3)/(6x(x+1))`

`dy/dx=(e^(x^2-x)(12x^3+6x^2+x+3))/(6sqrt(x)(x+1)^(1/3))`

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

Taking the natural logarithm of both sides and applying the properties of logarithm, we get

`logy=logsqrt(x)+loge^((x^2-x)(x+1)^(2/3))`

`logy=1/2(logx)+(x^2-x)(x+1)^(2/3)loge`

`logy=1/2(logx)+(x^2-x)(x+1)^(2/3)`

Differentiating both sides with respect to x, we get

`1/y dy/(dx)=1/(2x)+(x^2-x) d/(dx) (x+1)^(2/3) +(x+1)^(2/3) d/(dx) (x^2-x)`

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

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

`dy/dx=(sqrt(x)e^((x^2-x)(x+1)^(2/3)))((3(x+1)^(1/3)+4x(x^2-x)+6x(2x-1)(x+1)))/(6x(x+1)^(1/3))`

`dy/dx=(sqrt(x)e^((x^2-x)(x+1)^(2/3)))(((3(x+1)^(1/3)+4x^3-4x^2+12x^3+12x^2-6x^2-6x))/(6x(x+1)^(1/3)))`

`dy/dx=(sqrt(x)e^((x^2-x)(x+1)^(2/3)))((3(x+1)^(1/3)+16x^3+2x^2-6x))/(6x(x+1)^(1/3))`

`dy/dx=(e^((x^2-x)(x+1)^(2/3))(3(x+1)^(1/3)+16x^3+2x^2-6x))/(6sqrt(x)(x+1)^(1/3))`

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Correct answer posted in answer 2 , as there was mistake in noting down the question.