If y = tan x tan 2x tan 3x,then frac{dy}{dx} | vedclass

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(D) Given $y = 	an x 	an 2x 	an 3x$. We know that $	an 3x = 	an(2x + x) = \frac{	an 2x + 	an x}{1 - 	an x 	an 2x}$. Rearranging this,we get $

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All of the above

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(D) Given $y = 	an x 	an 2x 	an 3x$. We know that $	an 3x = 	an(2x + x) = \frac{	an 2x + 	an x}{1 - 	an x 	an 2x}$. Rearranging this,we get $	an 3x(1 - 	an x 	an 2x) = 	an 2x + 	an x$. $	an 3x - 	an x 	an 2x 	an 3x = 	an 2x + 	an x$. Thus,$y = 	an x 	an 2x 	an 3x = 	an 3x - 	an 2x - 	an x$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(	an 3x) - \frac{d}{dx}(	an 2x) - \frac{d}{dx}(	an x)$. $\frac{dy}{dx} = 3 \sec^2 3x - 2 \sec^2 2x - \sec^2 x$. This matches option $C$. Since the expression can be manipulated into other forms,let us check if $A$ and $B$ are equivalent. Using logarithmic differentiation,$\ln y = \ln(	an x) + \ln(	an 2x) + \ln(	an 3x)$. $\frac{1}{y} \frac{dy}{dx} = \frac{\sec^2 x}{	an x} + \frac{2 \sec^2 2x}{	an 2x} + \frac{3 \sec^2 3x}{	an 3x}$. $\frac{dy}{dx} = y (\frac{1}{\sin x \cos x} + \frac{2}{\sin 2x \cos 2x} + \frac{3}{\sin 3x \cos 3x}) = y (2 \csc 2x + 4 \csc 4x + 6 \csc 6x) = 2y (\csc 2x + 2 \csc 4x + 3 \csc 6x)$. This matches option $B$. Since both $B$ and $C$ are correct,option $D$ is the correct choice.

If $y = \tan x \tan 2x \tan 3x$,then $\frac{dy}{dx}$ is equal to:

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