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

Question

4587 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-25T21:09:34

#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 see

#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)#

Answer 2 · 2018-03-25T21:17:41

#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)#