Question

Differentiate (x² – 5x + 8) (x³ + 7x + 9) in three ways mentioned below:

1. by using product rule
2. by expanding the product to obtain a single polynomial.
3. by logarithmic differentiation.

Do they all give the same answer?

Answer 1

(i) By using the product rule.

Let, y = (x² – 5x + 8). (x³ + 7x + 9)

Differentiating both sides with respect to x,

`dy/dx = (x^2 = 5x + 8) d/dx(x^3 + 7x + 9) + (x^3 + 7x + 9) d/dx (x^2 - 5x + 8)`

`dy/dx = (x^2 - 5x + 8) (3x^2 + 7) + (x^3 + 7x + 9)(2x - 5)`

`= 3x^2 (x^2 - 5x + 8) + 7 (x^2 - 5x + 8) + 2x (x^3 + 7x + 9) - 5 (x^3 + 7x + 9)`

`= 3x^4 - 15x^3 + 24x^2 + 7x^2 - 35x + 56 + 2x^4 + 14x^2 + 18 x - 5x^3 - 35 x - 45`       ...(1)

`= 5x^4 - 20x^3 + 45x^2 - 52x + 11`    ....(1)

(ii) By expanding the product to obtain a single polynomial

y = (x² – 5x + 8) (x³ + 7x + 9)

`= x^2 (x^3 + 7x + 9) - 5x (x^3 + 7x + 9) + 8 (x^3 + 7x + 9)`

`= x^5 + 7x^3 + 9x^2 - 5x^4 - 35x^2 - 45x + 8x^3 + 56x + 72`

`= x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72`

Differentiating both sides with respect to x,

`dy/dx = 5x^4 - 20x^3 + 45x^2 - 52x + 11`          ...(2)

(iii) By logarithmic differentiation.

Let, y = (x² – 5x + 8) (x³ + 7x + 9) 

Taking logarithm of both sides,

log y = log (x² – 5x + 8) + log (x³ + 7x + 9)             ...[∵ log (mn) = log m + log n]

Differentiating both sides with respect to x,

`1/y dy/dx = 1/(x^2 - 5x + 8) d/dx (x^2 - 5x + 8) + 1/(x^3 + 7x + 9) d/dx (x^3 + 7x + 9)`

`= (2x - 5)/(x^2 - 5x + 8) + (3x^2 + 7)/(x^3 + 7x + 9)`

`= ((2x - 5)(x^3 + 7x + 9) + (3x^2 + 7) (x^2 - 5x + 8))/((x^2 - 5x + 8)(x^3 + 7x + 9))`

`therefore dy/dx = y [(2x (x^3 + 7x + 9) - 5 (x^3 + 7x + 9) + 3x^2 (x^2 - 5x + 8) + 7 (x^2 - 5x + 8))/((x^2 - 5x + 8)(x^3 + 7x + 9))]`

`= (x^2 - 5x + 8) (x^3 + 7x + 9) [(2x^4 + 14x^2 + 18x - 5x^3 - 35x - 45 + 3x^4 - 15x^3 + 24x^2 + 7x^2 - 35x + 56)/((x^2 - 5x + 8)(x^3 + 7x + 9))]`

= 5x⁴ - 20x³ + 45x² - 52x + 11           ...(3)

It is clear from equations (1), (2) and (3) that the values ​​of `dy/dx` are equal.