Question

Use logarithmic differentiation to find dy/dx if y= sqrt of(x^3-1)(x^3+2)?

Answer 1

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

Explanation:

Let:

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

using the properties of logarithms:

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

and differentiating:

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

simplifying:

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

`(dg)/dx = 1/2(3x^5+6x^3 + 3x^5+3x^3)/((x^3+2)(x^3-1))`

`(dg)/dx = 1/2(6x^5+9x^3 )/((x^3+2)(x^3-1))`

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

Consider now that using the chain rule:

`(dg)/dx = (dg)/(dy)dy/dx = d/dy(lny)dy/dx = 1/y dy/dx`

and then:

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

and finally:

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