Differentiate the given functions w.r.t. x.

cos x cos 2x cos 3x

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Solution

Let y = cos x cos 2x cos 3x

Taking log on both side, we get

log y = log (cos x cos 2x cos 3x) [log m $\times n \times$ 1=log m+log n+log l ]

or log y = log (cos x)+log (cos 2x) + log (cos 3x)

Differentiating both sides w.r.t. x, we get

$\frac{d}{d x} l o g y = \frac{d}{d x} l o g (c o s x) + \frac{d}{d x} l o g (c o s 2 x) + \frac{d}{d x} l o g (c o s 3 x) \frac{1}{y} \frac{d y}{d x} = \frac{1}{c o s x} (- s i n x) + \frac{1}{c o s 2 x} (- s i n 2 x \frac{d}{d x} (2 x)) + \frac{1}{c o s 3 x} (- s i n 3 x . \frac{d}{d x} (3 x)) \Rightarrow \frac{d y}{d x} = y {- t a n x - 2 t a n 2 x - 3 t a n 3 x} = - y {t a n x + 2 t a n 2 x + 3 t a n 3 x} = - c o s x c o s 2 x c o s 3 x {t a n x + 2 t a n 2 x + 3 t a n 3 x}$

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