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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following w.r.t. x:
$(\text{x}+1)^2(\text{x}+2)^3(\text{x}+3)^4$
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Answer

Let $\text{y}=(\text{x}+1)^2(\text{x}+2)^3(\text{x}+3)^4$
$\therefore\ \log\text{y}=\log\big\{(\text{x}+1)^2\cdot(\text{x}+2)^3(\text{x}+3)^4\big\}$
$=\log(\text{x}+1)^2+\log(\text{x}+2)^3+\log(\text{x}+3)^4$
and $\frac{\text{d}}{\text{dy}}\log\text{y}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[2\log(\text{x}+1)\big]\\+\frac{\text{d}}{\text{dx}}\big[3\log(\text{x}+2)\big]+\frac{\text{d}}{\text{dx}}\big[4\log(\text{x}+3)\big]$
$\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\frac{2}{(\text{x}+1)}\cdot\frac{\text{d}}{\text{dx}}(\text{x}+1)+3\cdot\frac{1}{(\text{x}+2)}\cdot\\\frac{\text{d}}{\text{dx}}(\text{x}+2)+4\cdot\frac{1}{(\text{x}+3)}\cdot\frac{\text{d}}{\text{dx}}(\text{x}+3)$$\Big[\because\frac{\text{d}}{\text{dx}}(\log\text{x})=\frac{1}{\text{x}}\Big]$
$=\Big[\frac{2}{\text{x}+1}+\frac{3}{\text{x}+2}+\frac{4}{\text{x}+3}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{2}{\text{x}+1}+\frac{3}{\text{x}+2}+\frac{4}{\text{x}+3}\Big]$
$=(\text{x}+1)^2\cdot(\text{x}+2)^3\cdot(\text{x}+3)^4\Big[\frac{2}{\text{x}+1}+\frac{3}{\text{x}+2}+\frac{4}{\text{x}+3}\Big]$
$=(\text{x}+1)^2\cdot(\text{x}+2)^3\cdot(\text{x}+3)^4$
$\bigg[\frac{2(\text{x}+2)(\text{x}+3)+3(\text{x}+1)(\text{x}+3)+4(\text{x}+1)(\text{x}+2)}{(\text{x}+1)(\text{x}+2)(\text{x}+3)}\bigg]$
$=\frac{(\text{x}+1)^2(\text{x}+2)^3(\text{x}+3)^4}{(\text{x}+1)(\text{x}+2)(\text{x}+3)}$
$\big[2(\text{x}^2+5\text{x}+6)+3(\text{x}^2+4\text{x}+3)+4(\text{x}^2+3\text{x}+2)\big]$
$=(\text{x}+1)(\text{x}+2)^2(\text{x}+3)^3$
$\big[2\text{x}^2+10\text{x}+12+3\text{x}^2+12\text{x}+9+4\text{x}^2+12\text{x}+8\big]$
$=(\text{x}+1)(\text{x}+2)^2(\text{x}+3)^3\big[9\text{x}^2+34\text{x}+29\big]$

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