What is the derivative of #f(x)=cos^2x*cos2x#?

Question

Emily Ammerman · Accepted Answer

#f'(x)=-2sin(2x)cos^2(x)-2cos(2x)sin(2x)#

Recall that the derivative of two multiplied functions, #f(x)=a(x)b(x),# is given by #f'(x)=a(x)b'(x)+b(x)a'(x)#

Here, we observe

#a(x)=cos^2(x)#

#a'(x)=2cos(x)*d/dxcos(x)=-2cosxsinx=-sin(2x)# (From the identity #sin(2x)=2sinxcosx#)

#b(x)=cos(2x)#

#b'(x)=-sin(2x)*d/dx(2x)=-2sin(2x)#

Thus,

#f'(x)=-2sin(2x)cos^2(x)-2cos(2x)sin(2x)#

Isabella Ackerman · Answer

#f'(x)=-8cos^3(x)sin(x)+2cos(x)sin(x)#

We have #f(x)=cos^2(x)cos(2x)#

Let's use the trigonometric identity:

#cos^2(x)=(1+cos(2x))/2rArrcos(2x)=2cos^2(x)-1#

so we can express the trig functions in terms of a common angle #x#

#rArrf(x)=cos^2(x)(2cos^2(x)-1)#

#rArrf(x)=2cos^4(x)-cos^2(x)#

Now we can take the derivative using the chain rule:

#rArrf'(x)=8cos^3(x)(-sin(x))-2cos(x)(-sin(x))#

#rArrf'(x)=-8cos^3(x)sin(x)+2cos(x)sin(x)#

Isaiah Allen · Answer

undefined To find the derivative of $ f(x) = \cos^2(x) \cdot \cos(2x) $, you can use the product rule.

The product rule states that if $ u(x) $ and $ v(x) $ are differentiable functions, then the derivative of their product $ u(x) \cdot v(x) $ with respect to $ x $ is $ u'(x) \cdot v(x) + u(x) \cdot v'(x) $.

Apply the product rule:

$$ f'(x) = (\cos^2(x))' \cdot \cos(2x) + \cos^2(x) \cdot (\cos(2x))' $$

Find the derivatives of $ \cos^2(x) $ and $ \cos(2x) $:

$$ (\cos^2(x))' = 2\cos(x) \cdot (-\sin(x)) = -2\cos(x) \sin(x) $$

$$ (\cos(2x))' = -2\sin(2x) $$

Plug these derivatives back into the product rule formula:

$$ f'(x) = -2\cos(x) \sin(x) \cdot \cos(2x) + \cos^2(x) \cdot (-2\sin(2x)) $$

$$ f'(x) = -2\cos(x) \sin(x) \cos(2x) - 2\cos^2(x) \sin(2x) $$

So, the derivative of $ f(x) = \cos^2(x) \cdot \cos(2x) $ is $ f'(x) = -2\cos(x) \sin(x) \cos(2x) - 2\cos^2(x) \sin(2x) $.