Differentiate

Question:

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

Solution:

Let $y=\left(x^{5}-5 x+8\right)\left(x^{3}+7 x+9\right)$

(i)

Let $x^{2}-5 x+8=u$ and $x^{3}+7 x+9=v$

$\therefore y=u v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x} \quad$ (By using product rule)

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^{2}-5 x+8\right) \cdot\left(x^{3}+7 x+9\right)+\left(x^{2}-5 x+8\right) \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right)$

$\Rightarrow \frac{d y}{d x}=(2 x-5)\left(x^{3}+7 x+9\right)+\left(x^{2}-5 x+8\right)\left(3 x^{2}+7\right)$

$\Rightarrow \frac{d y}{d x}=2 x\left(x^{3}+7 x+9\right)-5\left(x^{3}+7 x+9\right)+x^{2}\left(3 x^{2}+7\right)-5 x\left(3 x^{2}+7\right)+8\left(3 x^{2}+7\right)$

$\Rightarrow \frac{d y}{d x}=\left(2 x^{4}+14 x^{2}+18 x\right)-5 x^{3}-35 x-45+\left(3 x^{4}+7 x^{2}\right)-15 x^{3}-35 x+24 x^{2}+56$

$\therefore \frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11$

(ii)

\begin{aligned} y &=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \\ &=x^{2}\left(x^{3}+7 x+9\right)-5 x\left(x^{3}+7 x+9\right)+8\left(x^{3}+7 x+9\right) \\ &=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 \end{aligned}

$\therefore \frac{d y}{d x}=\frac{d}{d x}\left(x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72\right)$

$=\frac{d}{d x}\left(x^{5}\right)-5 \frac{d}{d x}\left(x^{4}\right)+15 \frac{d}{d x}\left(x^{3}\right)-26 \frac{d}{d x}\left(x^{2}\right)+11 \frac{d}{d x}(x)+\frac{d}{d x}(72)$

$=5 x^{4}-5 \times 4 x^{3}+15 \times 3 x^{2}-26 \times 2 x+11 \times 1+0$

$=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11$

(iii) $y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$

Taking logarithm on both the sides, we obtain

$\log y=\log \left(x^{2}-5 x+8\right)+\log \left(x^{3}+7 x+9\right)$

Differentiating both sides with respect to x, we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x} \log \left(x^{2}-5 x+8\right)+\frac{d}{d x} \log \left(x^{3}+7 x+9\right)$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}-5 x+8} \cdot \frac{d}{d x}\left(x^{2}-5 x+8\right)+\frac{1}{x^{3}+7 x+9} \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right)$

$\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x^{2}-5 x+8} \times(2 x-5)+\frac{1}{x^{3}+7 x+9} \times\left(3 x^{2}+7\right)\right]$

$\Rightarrow \frac{d y}{d x}=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\left[\frac{2 x-5}{x^{2}-5 x+8}+\frac{3 x^{2}+7}{x^{3}+7 x+9}\right]$

$\Rightarrow \frac{d y}{d x}=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\left[\frac{(2 x-5)\left(x^{3}+7 x+9\right)+\left(3 x^{2}+7\right)\left(x^{2}-5 x+8\right)}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$

$\Rightarrow \frac{d y}{d x}=2 x\left(x^{3}+7 x+9\right)-5\left(x^{3}+7 x+9\right)+3 x^{2}\left(x^{2}-5 x+8\right)+7\left(x^{2}-5 x+8\right)$

$\Rightarrow \frac{d y}{d x}=\left(2 x^{4}+14 x^{2}+18 x\right)-5 x^{3}-35 x-45+\left(3 x^{4}-15 x^{3}+24 x^{2}\right)+\left(7 x^{2}-35 x+56\right)$

$\Rightarrow \frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11$

From the above three observations, it can be concluded that all the results of $\frac{d y}{d x}$ are same.