Question

Differentiate the following w.r.t. $x$:
$cosx.cos2x.cos3x$

Answer 1

The correct answer is $∴\frac{dy}{dx}=-cos\,x.cos\,2x.cos\,3x[tan\,x+2tan\,2x+3tan\,3x]$
Let $y=cosx.cos2x.cos3x$
Taking logarithm on both the sides,we obtain
$log\,y=log(cos\,x.cos\,2x.cos\,3x)$
$⇒log\,y=log(cos\,x)+log(cos\,2x)+log(cos\,3x)$
Differentiating both sides with respect to x,we obtain
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{cos\,x}.\frac{d}{dx}(cos\,x)+\frac{1}{cos\,2x}.\frac{d}{dx}(cos\,2x)+\frac{1}{cos\,3x}.\frac{d}{dx}(cos\,3x)$
$⇒\frac{dy}{dx}=y[-\frac{sin\,x}{cos\,x}-\frac{sin\,2x}{cos\,2x}.\frac{d}{dx}(2x)-\frac{sin\,3x}{cos\,3x}.\frac{d}{dx}(3x)]$
$∴\frac{dy}{dx}=-cos\,x.cos\,2x.cos\,3x[tan\,x+2tan\,2x+3tan\,3x]$