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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate the function given in Exercise:
$\cos\text{x}.\cos 2\text{x}.\cos 3\text{x}$
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Answer

Let $\text{y}=\cos\text{x}\cos2\text{x}\cos3\text{x}\ \dots\text{(i})$
Taking logs on both sides, we have
$\log\text{y}=\log(\cos\text{x}\cos2\text{x}\cos3\text{x})$ $=\log\cos\text{x}+\log\cos2\text{x}+\log\cos3\text{x}$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\frac{\text{d}}{\text{dx}}\log\cos2\text{x}+\frac{\text{d}}{\text{dx}}\log\cos3\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{d}}{\text{dx}}=\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}\cos\text{x}+\frac{1}{\cos2\text{x}}\frac{\text{d}}{\text{dx}}\cos2\text{x}+\frac{1}{\cos3\text{x}}\frac{\text{d}}{\text{dx}}\cos3\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{d}}{\text{dx}}=\frac{1}{\cos\text{x}}(-\sin\text{x})+\frac{1}{\cos2\text{x}}(-\sin2\text{x})\frac{\text{d}}{\text{dx}}2\text{x}+\frac{1}{\cos3\text{x}}(-\sin3\text{x})\frac{\text{d}}{\text{dx}}3\text{x}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=-\tan\text{x}-(\tan2\text{x})2-\tan3\text{x}(3)$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}=-\text{y}(\tan\text{x}+2\tan2\text{x}+3\tan3\text{x})$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}=-\cos\text{x}\cos2\text{x}\cos3\text{x}(\tan\text{x}+2\tan2\text{x}+3\tan3\text{x})\ \ [\text{From eq. (i)}]$

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