Differentiate `(x^2-5x+8)(x^3+7x+9)` in three ways mentioned below:(i) by using product rule(ii) by expanding the product to obtain a single polynomial.(iii) by logarithmic differentiation.Do they all give the same answer?

Explanation

`"Let "y = (x^(2) -5x + 8) (x^(3) + 7x +9)`
`(i) (dy)/(dx) = (x^(2) -5x + 8) (d)/(dx) (x^(3) + 7x +9) + (x^(3) + 7x +9) (d)/(dx) (x^(2) -5x + 8)`
`=(x^(2) - 5x + 8)(3x^(2) + 7) + (x^(3) + 7x + 9)(2x - 5)`
`= 3x^(4) + 7x^(2) - 15x^(3) - 35x + 24x^(2) + 56 + 2x^(4) - 5x^(3) + 14x^(2) - 35x + 18x -45`
`= 5x^(4) - 20x^(3) + 45x^(2) - 52x + 11`
`(ii) y = (x^(2) - 5x + 8)(x^(3) + 7x + 9)`
`= (x^(5) - 5x^(4) + 15x^(3) - 26x^(2) + 11x +72)`
`rArr (dy)/(dx)= (d)/(dx) (x^(5) - 5x^(4) + 15x^(3) - 26x^(2) + 11x +72)`
`= 5x^(4) - 20x^(3) + 45x^(2) - 52x + 11`
`(iii) y= (x^(2) - 5x + 8)(x^(3) + 7x + 9)`
`rArr "log"y = "log" [(x^(2) - 5x + 8) * (x^(3) + 7x + 9)]`
`= "log" (x^(2) - 5x + 8) +"log" (x^(3) + 7x + 9)`
Differentiate both side w.r.t.x
`(1)/(y) (dy)/(dx) = (2x - 5)/(x^(2) - 5x + 8) + (3x^(2) + 7)/(x^(3) + 7x + 9)`
`rArr (dy)/(dx) = y[((2x - 5)(x^(3) + 7x + 9) + (3x^(2) + 7)(x^(2) - 5x + 8))/((x^(2) - 5x + 8)(x^(3) + 7x + 9))]`
`= y[(2x^(4) - 5x^(3) + 14x^(2) - 35x + 18x - 45 + 3x^(4) + 7x^(2) - 15x^(3) - 35x + 24x^(2) + 56)/(y)]`
`= 5x^(4) - 20x^(3) + 45x^(2) - 52x + 11`
Thus, the results in three cases are same.