Differentiate $(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$ 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?

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Solution

Let $y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$
(i) Let $x^{2} - 5 x + 8 = u$ and $x^{3} + 7 x + 9 = v$
$∴ y = u v$
$\Rightarrow \frac{d y}{d x} = \frac{d u}{d x} . v + u . \frac{d v}{d x}$
By using product rule,
$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (x^{2} - 5 x + 8) . (x^{3} + 7 x + 9) + (x^{2} - 5 x + 8) . \frac{d}{d x} (x^{3} + 7 x + 9)$
$\Rightarrow \frac{d y}{d x} = (2 x - 5) (x^{3} + 7 x + 9) + (x^{2} - 5 x + 8) (3 x^{2} + 7)$
$\Rightarrow \frac{d y}{d x} = 2 x (x^{3} + 7 x + 9) - 5 (x^{3} + 7 x + 9) + x^{2} (3 x^{2} + 7) - 5 x (3 x^{2} + 7) + 8 (3 x^{2} + 7)$
$\Rightarrow \frac{d y}{d x} = (2 x^{4} + 14 x^{2} + 18 x) - 5 x^{3} - 35 x - 45 + (3 x^{4} + 7 x^{2}) - 15 x^{3} - 35 x + 24 x^{2} + 56$
$∴ \frac{d y}{d x} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11$
(ii) $y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$
$= x^{2} (x^{3} + 7 x + 9) - 5 x (x^{3} + 7 x + 9) + 8 (x^{3} + 7 x + 9)$
$= x^{5} + 7 x^{3} + 9 x^{2} - 5 x^{4} - 35 x^{2} - 45 x + 8 x^{3} + 56 x + 72$
$= x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72$
$∴ \frac{d y}{d x} = \frac{d}{d x} (x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72)$
$= \frac{d}{d x} (x^{5}) - 5 \frac{d}{d x} (x^{4}) + 15 \frac{d}{d x} (x^{3}) - 26 \frac{d}{d x} (x^{2}) + 11 \frac{d}{d x} (x) + \frac{d}{d x} (72)$
$= 5 x^{4} - 5 \times 4 x^{3} + 15 \times 3 x^{2} - 26 \times 2 x + 11 \times 1 + 0$
$= 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11$
(iii) $y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$
Taking logarithm on both the sides, we obtain
$log y = log (x^{2} - 5 x + 8) + log (x^{3} + 7 x + 9)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{y} \frac{d y}{d x} = \frac{d}{d x} log (x^{2} - 5 x + 8) + \frac{d}{d x} log (x^{3} + 7 x + 9)$
$\Rightarrow \frac{1}{y} \frac{d y}{d x} = \frac{1}{x^{2} - 5 x + 8} . \frac{d}{d x} (x^{2} - 5 x + 8) + \frac{1}{x^{3} + 7 x + 9} . \frac{d}{d x} (x^{3} + 7 x + 9)$
$\Rightarrow \frac{d y}{d x} = y [\frac{1}{x^{2} - 5 x + 8} \times (2 x - 5) + \frac{1}{x^{3} + 7 x + 9} \times (3 x^{2} + 7)]$
$\Rightarrow \frac{d y}{d x} = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) [\frac{2 x - 5}{x^{2} - 5 x + 8} + \frac{3 x^{2} + 7}{x^{3} + 7 x + 9}]$
$\Rightarrow \frac{d y}{d x} = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) [\frac{(2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)}{(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)}]$
$\Rightarrow \frac{d y}{d x} = 2 x (x^{3} + 7 x + 9) - 5 (x^{3} + 7 x + 9) + 3 x^{2} (x^{2} - 5 x + 8) + 7 (x^{2} - 5 x + 8)$
$\Rightarrow \frac{d y}{d x} = (2 x^{4} + 14 x^{2} + 18 x) - 5 x^{3} - 35 x - 45 + (3 x^{4} - 15 x^{3} + 24 x^{2}) + (7 x^{2} - 35 x + 56)$
$\Rightarrow \frac{d y}{d x} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11$
From the above three observations, it can be concluded that all the $\frac{d y}{d x}$ results of are same.

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Similar questions

Differentiate 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?

Differentiate $(x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$ in three ways mentioned below.

(a) By using product rule.

(b) By expanding the product to obtain a single polynomial.

Do they all given the same answer?

\begin{matrix} Differentiate (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) in ways mentioned below: (i) by using product rule \end{matrix}

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