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Differentiate $ (x^2 - 5x + 8) (x^3 + 7x + 9 ) $ in three ways mentioned below:\begin{array}{1 1} (i)\;by\;using\;product\;rule & \;\\(ii)\;by\;expanding\;the\;product\;to\;obtain\;a\;single\;polynomial. & \;\\ (iii)\;by\;logarithmic\;differentiation. & \;\end{array} Do they all give the same answer?

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Toolbox:
  • The product rule is $(uv)'=u'v+uv'$
  • $\log mn=\log m+\log n$
Step 1:
We have $y=(x^2-5x+8)(x^3+7x+9)$
(i) Differentiating by product rule
$\large\frac{dy}{dx}=[\large\frac{d}{dx}$$(x^2-5x+8)](x^3+7x+9)+(x^2-5x+8)\large\frac{d}{dx}$$(x^3+7x+9)$
$\quad\;\;=(2x-5)(x^3+7x+9)+(x^2-5x+8)(3x^2+7)$
$\quad\;\;=2x(x^3+7x+9)-5(x^3+7x+9)+3x^2(x^2-5x+8)+7(x^2-5x+8)$
$\Rightarrow 2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56$
$\Rightarrow 5x^4-20x^3+45x^2-52x+11$
$\large\frac{dy}{dx}=$$5x^4-20x^3+45x^2-52x+11$
Step 2:
(ii) By expanding the product to obtain a single polynomial.
$y=(x^2-5x+8)(x^3+7x+9)$
$\;\;=x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)$
$\;\;=x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72$
$\large\frac{dy}{dx}=$$5x^4+21x^2+18x-20x^3-70x-45+24x^2+56+0$
$\;\quad=5x^4-20x^3+45x^2-52x+11.$
Step 3:
(iii) By lograthimic differentiation
$\log mn=\log m+\log n$
$y=(x^2-5x+8)(x^3+7x+9)$
Taking $\log$ on both sides
$\log y=\log(x^2-5x+8)\log(x^3+7x+9)$
$\;\;\quad\;\;=\log(x^2-5x+8)+\log(x^3+7x+9)$
Differentiating with respect to $x$
$\large\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2-5x+8}\frac{d}{dx}$$(x^2-5x+8)+\large\frac{1}{x^3+7x+9}\frac{d}{dx}$$(x^3+7x+9)$
$\qquad=\large\frac{1}{x^2-5x+8}$$(2x-5)+\large\frac{1}{x^3+7x+9}$$(3x^2+7)$
$\qquad\;=\large\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}$
$\large\frac{dy}{dx}=$$y\bigg(\large\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}\bigg)$
$\large\frac{dy}{dx}=$$(x^2-5x+8)(x^2+7x+9)\bigg(\large\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}\bigg)$
$\quad\;=(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)$
$\quad\;=2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56$
$\quad\;=5x^4-20x^3+45x^2-52x+11$
Step 4:
Hence the answer is the same in all the three cases.
answered May 9, 2013 by sreemathi.v
edited May 10, 2013 by sreemathi.v
 

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