What is the derivative of $f (x) = {cos}^{2} x \cdot cos 2 x$ $f(x)=cos^2x*cos2x$ ?

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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:

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

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

What is the derivative of f(x)=cos^2x*cos2xf(x)=cos2x⋅cos2xf(x)=cos^2x*cos2x?