Question

Differentiate $(x^5-5x+8)(x^3+7x+9)$ in three ways mentioned below
(i) By using product rule
(ii) By expanding the product to obtain a single polynomial.
(iii) By logarithmic differentiation.
Do they all give the same answer?

Answer 1

(i)Let $y=(x^5-5x+8)(x^3+7x+9)$
let $x^5-5x+8=u$ and $x^3+7x+9=v$
$∴y=uv$
$⇒\frac{dy}{dx}=\frac{du}{dx}.v+u.\frac{dv}{dx}$ [by product rule]
$⇒\frac{dy}{dx}=\frac{d}{dx}(x^5-5x+8).(x^3+7x+9)+(x^5-5x+8).\frac{d}{dx}(x^3+7x+9)$
$⇒\frac{dy}{dx}=(2x-5)(x^3+7x+9)+(x^5-5x+8)(3x^2+7)$
$⇒\frac{dy}{dx}=2x(x^3+7x+9)-5(x^3+7x+9)+x^2(3x^2+7)-5x(3x^2+7)+8(3x^2+7)$
$⇒\frac{dy}{dx}=(2x^4+14x^2+18x)-5x^3-35x-45+(3x^4+7x^2)-15x^3-35x+24x^2+56$
$∴\frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11$

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(ii)$y=(x^5-5x+8)(x^3+7x+9)$
$=x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)$
$=x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72$
$=x^5-5x^4+15x^3-26x^2+11x+72$
$∴\frac{dy}{dx}=\frac{d}{dx}(x^5-5x^4+15x^3-26x^2+11x+72)$
$=\frac{d}{dx}(x^5)-5\frac{d}{dx}(x^4)+15\frac{d}{dx}(x^3)-26\frac{d}{dx}(x^2)+11\frac{d}{dx}(x)+\frac{d}{dx}(72)$
$=5x^4-5\times4x^3+15\times3x^2-26\times2x+11\times1+0$
$=5x^4-20x^3+45x^2-52x+11$

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(iii)$y=(x^2-5x+8)(x^3+7x+9)$
Taking logarithm on both the sides,we obtain
$logy=log(x^2-5x+8)+log(x^3+7x+9)$
Differentiating both sides with respect to $x$,we obtain
$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}log(x^2-5x+8)+\frac{d}{dx}log(x^3+7x+9)$
$⇒\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^5-5x+8}.\frac{d}{dx}(x^2-5x+8)+\frac{1}{x^3+7x+9}.\frac{d}{dx}(x^3+7x+9)$
$⇒\frac{dy}{dx}=y[\frac{1}{x^2-5x+8}.(2x-5)+\frac{1}{x^3+7x+9}.(3x^2+7)]$
$⇒\frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9)[\frac{2x-5}{(x^2-5x+8)}+\frac{3x^2+7}{(x^3+7x+9)}]$
$⇒\frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9)[\frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)}]$
$⇒\frac{dy}{dx}=2x(x^3+7x+9)-5(x^3+7x+9)+3x^2(x^2-5x+8)+7(x^2-5x+8)$
$⇒\frac{dy}{dx}=(2x^4+14x^2+18x)-5x^3-35x-45+(3x^4-15x^3+24x^2)+(7x^2-35x+56)$
$⇒\frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11$
From the above three observations, it can be concluded that all the results of dy/dx are same