Question

Y=cos4xcos2x , find dy/dx

Answer 1

ANSWER

The required derivative of y or dy/dx is :

-2(cos 4x sin 2x + 2cos 2x sin 4x)

GIVEN

• y = cos4x.cos2x

TO FIND

• The derivative of y with respect to x i.e. dy/dx

SOLUTION

$\rm \implies y = \cos4x \cos2x \\ \\ \rm \implies \dfrac{dy}{dx} = \dfrac{d}{dx} \{ \cos4x \cos2x \} \\ \\ \rm \implies \dfrac{dy}{dx} = \cos4x \dfrac{d}{dx} ( \cos2x) + \cos2x \dfrac{d}{dx} ( \cos4x) \\ \\ \rm \implies \dfrac{dy}{dx} = \cos4x( - \sin2x) \dfrac{d}{dx} (2x) + \cos2x( - \sin4x) \dfrac{d}{dx} (4x) \\ \\ \rm \implies \dfrac{dy}{dx} = - 2 \cos4x \sin2x - 4 \cos2x \sin4x \\ \\ \rm \implies \frac{dy}{dx} = - 2( \cos4x \sin2x + 2 \cos2x \sin4x)$

Answer 2

Given ,

The function is Y = Cos(4x) × Cos(2x)

We know that , the product rule is given by

$\large \sf \fbox{ \frac{d(u.v)}{dx} =v \frac{du}{dx} + u \frac{dv}{dx} }$

Thus ,

$\sf \mapsto \frac{dy}{dx} = Cos(2x)\frac{Cos(4x)}{dx} + Cos(4x) \frac{Cos(2x)}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} = - Cos(2x). Sin(4x) \frac{d(4x)}{dx} + ( - Cos(4x) .Sin(2x) \frac{d(2x)}{x} ) \\ \\ \sf \mapsto \frac{dy}{dx} = - 4Cos(2x). Sin(4x) - 2Cos(4x) .Sin(2x)$

Remmember :

$\sf \mapsto \frac{d(Cos(x) )}{dx} = -Sin(x) \\ \\ \sf \mapsto \frac{dx}{dx} = 1 \\ \\ \sf \mapsto \frac{d(Constant)}{dx} = 0$