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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}$
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Answer

We have, $\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}.....(\text{i})$$\text{y}=\frac{(\text{x}^2-1)^3(2\text{x}-1)}{(\text{x}-3)^{\frac{1}{2}}(4\text{x}-1)^{\frac{1}{2}}}$
Taking log on both sides, $\log\text{y}=\log\Bigg[\frac{(\text{x}^2-1)^3(2\text{x}-1)}{(\text{x}-3)^{\frac{1}{2}}(4\text{x}-1)^{\frac{1}{2}}}\Bigg]$$\Rightarrow\log\text{y}=\log(\text{x}^2-1)^3+\log(2\text{x}-1)-\log(\text{x}-3)^{\frac{1}{2}}-\log(4\text{x}-1)^{\frac{1}{2}}$
$\Rightarrow\log\text{y}=\log(\text{x}^2-1)^3+\log(2\text{x}-1)-\frac{1}{2}\log(\text{x}-3)-\frac{1}{2}\log(4\text{x}-1)$
Differentiating with respect to x using chain rule, $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\frac{\text{d}}{\text{dx}}\Big\{\log(\text{x}^2-1)\Big\}+\frac{\text{d}}{\text{dx}}\Big\{\log(2\text{x}-1)\Big\}\\-\frac{1}{2}\frac{\text{d}}{\text{dx}}\Big\{\log(\text{x}-3)\Big\}-\frac{1}{2}\Big\{\log(4\text{x}-1)\Big\}$$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{1}{\text{x}^2-1}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)+\frac{1}{(2\text{x}-1)}\frac{\text{d}}{\text{dx}}(2\text{x}-1)\\-\frac{1}{2}\Big(\frac{1}{\text{x}-3}\Big)\frac{\text{d}}{\text{dx}}(\text{x}-3)-\frac{1}{2}\frac{1}{(4\text{x}-1)}\frac{\text{d}}{\text{dx}}(4\text{x}-1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{1}{\text{x}^2-1}\Big)(2\text{x})+\frac{1}{2\text{x}-1}(2)-\frac{1}{2}\Big(\frac{1}{\text{x}-3}\Big)(1)-\frac{1}{2}\Big(\frac{1}{4\text{x}-1}\Big)(4)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
$\Rightarrow\frac{(\text{x}^2-1)^3(2\text{x}-1)}{\sqrt{(\text{x}-3)(4\text{x}-1)}}\Big[\frac{6\text{x}}{\text{x}^2-1}+\frac{2}{2\text{x}-1}-\frac{1}{2(\text{x}-3)}-\frac{2}{4\text{x}-1}\Big]$
[Using equation (i)]

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