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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability3 Marks
Question
Differentiate the function $\cos x \cdot \cos 2 x \cdot \cos 3 x$ w.r.t. x.
✓

Answer

Given function is: $cosx.cos2.cos3x$
Let y = $cosx.cos2.cos3x$
Taking log on both sides, we get
log y = log(cos x. cos 2x. cos 3x)
Now, differentiate both sides with respect to x
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})=\frac{\mathrm{d}}{\mathrm{dx}} \log (\cos \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}} \log (\cos 2 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(\log \cos 3 \mathrm{x})$ 
$\implies$ $\frac{1}{\mathrm{y}} \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{1}{\cos \mathrm{x}} \cdot \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\cos \mathrm{x})+\frac{1}{\cos 2 \mathrm{x}} \cdot \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\cos 2 \mathrm{x})+\frac{1}{\cos 3 \mathrm{x}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\cos 3 \mathrm{x})$ 
$\implies$ $\frac{d y}{d x}=y\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)\right]$
$\implies$ $\frac{\mathrm{dy}}{\mathrm{dx}}=-\cos \mathrm{x} \cdot \cos 2 \mathrm{x} \cdot \cos 3 \mathrm{x}[\tan \mathrm{x}+\tan 2 \mathrm{x}(2)+\tan 3 \mathrm{x}(3)]$
$\implies$ $\frac{\mathrm{dy}}{\mathrm{dx}}=-\cos \mathrm{x} \cdot \cos 2 \mathrm{x} \cdot \cos 3 \mathrm{x}[\tan \mathrm{x}+2 \tan 2 \mathrm{x}+3 \tan 3 \mathrm{x}]$

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