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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}$
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Answer

Here,
$\text{y}=\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x})$
$\log\text{y}=\log\sin\text{x}+\log\sin2\text{x}+\log\sin3\text{x}+\log\sin4\text{x}$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\frac{\text{d}}{\text{dx}}\log\sin3\text{x}+\frac{\text{d}}{\text{dx}}\log\sin4\text{x}$
$=\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\frac{1}{\sin3\text{x}}\frac{\text{d}}{\text{dx}}(\sin3\text{x})+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4 \text{x})$
$=\frac{1}{\sin\text{x}}(\cos\text{x})+\frac{1}{\sin2\text{x}}(\cos2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x}) \\ +\frac{1}{\sin3\text{x}}(\cos3\text{x})\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\big[\cot\text{x}+\cot2\text{x}(2)+\cot3\text{x}(3)+\cot4\text{x}(4)\big]$
$\frac{\text{dy}}{\text{dx}}=\text{y}\big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}) \\ \big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
[Using equation (i)]

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