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,

(iil) by logarithmic differentiation.
Do they all give the same answer?

(i) Let $f(x) = ({x^2} - 5x + 8)({x^3} + 7x + 9)$
By using product rule,

$f'(x) = {x^2} - 5x + 8)\cfrac{d}{{dx}}({x^3} + 7x + 9) + ({x^3} + 7x + 9)\cfrac{d}{{dx}}({x^2} - 5x + 8)$

$\Rightarrow$ $f'(x) = ({x^2} - 5x + 8)(3{x^2} + 7) + ({x^3} + 7x + 9)(2x - 5)$

$\Rightarrow$ $f'(x) = 3{x^4} - 15{x^3} + 24{x^2} + 7{x^2} - 35x + 56 + 2{x^4} + 14{x^2} + 18x - 5{x^3} - 35x - 45$
$\Rightarrow$ $f'(x) = 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11$

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

$f(x) = ({x^2} - 5x + 8)({x^3} + 7x + 9)$
$\Rightarrow$ $f(x) = {x^5} + 7{x^3} + 9{x^2} - 5{x^4} - 35{x^2} - 45x + 8{x^3} + 56x + 72$

$\Rightarrow$ $f(x) = {x^5} - 5{x^4} + 15{x^3} - 26{x^2} + 11x + 72$
therefore, $f'(x) = 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11$

(iii) By logarithmic differentiation,
Let $f(x) = y \Rightarrow y = ({x^2} - 5x + 8)({x^3} + 7x + 9)$

Taking log on both the sides, we get
$\log y = \log \{ ({x^2} - 5x + 8)({x^3} + 7x + 9)\}$
$\log y = \log \{ ({x^2} - 5x + 8) + \log ({x^3} + 7x + 9)\}$
Differentiating w.r.t. x, we get

$\cfrac{1}{y}\cfrac{{dy}}{{dx}} = \cfrac{1}{{({x^2} - 5x + 8)}}(2x - 5) + \cfrac{1}{{({x^3} + 7x + 9)}}(3{x^2} + 7)$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = y\left\{ {\cfrac{{2x - 5}}{{{x^2} - 5x + 8}} + \cfrac{{3{x^2} + 7}}{{{x^3} + 7x + 9}}} \right\}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = ({x^2} - 5x + 8)({x^3} + 7x + 9)\left[ {\cfrac{{2x - 5}}{{{x^2} - 5x + 8}} + \cfrac{{3{x^2} + 7}}{{{x^3} + 7x + 9}}} \right]$

$= (2x - 5)({x^3} + 7x + 9) + (3{x^2} + 7)({x^2} - 5x + 8)$
therefore, $\cfrac{{dy}}{{dx}} = 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11$
Yes, the answer is same in all the three cases.