Question

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

1. By using the 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)`

= (x² − 5x + 8) (3x² + 7) + (x³ + 7x + 9) (2x − 5)

= 3x² (x² − 5x + 8) + 7 (x² − 5x + 8) + 2x (x³ + 7x + 9) − 5 (x³ + 7x + 9)

= 3x⁴ − 15x³ + 24x² + 7x² − 35x + 56 + 2x⁴ + 14x² + 18x − 5x³ − 35x − 45

= 5x⁴ − 20x³ + 45x² − 52x + 11  ....(1)

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

y = (x² – 5x + 8) (x³ + 7x + 9)

= x² (x³ + 7x + 9) − 5x (x³ + 7x + 9) + 8 (x³ + 7x + 9)

= x⁵ + 7x³ + 9x² − 5x⁴ − 35x² − 45x + 8x³ + 56x + 72

= x⁵ − 5x⁴ + 15x³ − 26x² + 11x + 72

Differentiating both sides with respect to x,

`dy/dx` = 5x⁴ − 20x³ + 45x² − 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.