How do you find the derivative of $Cos^4(2x)$ ?

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Answer 1 · 2015-10-26T09:53:12

$-8sin(2x)cos^3(2x)$

You're in a case of the form $f^n(g(x))$ , where $f(x)=cos(x)$ , $g(x)=2x$ , and $n=4$ .

The rule for deriving power of functions states that

$[f^n(g(x))]' = n f^{n-1}(g(x)) * (f(g(x))'$

and the rule for composite function states that

$(f(g(x))'=f'(g(x)) * g'(x)$ .

So, the whole formula becomes

$[f^n(g(x))]' = n f^{n-1}(g(x)) * f'(g(x)) * g'(x)$

Let's analyse all the pieces: of course, if $n=4$ , $n-1=3$ .

Then, if $f(x)=cos(x)$ , $f'(x)=-sin(x)$ , and so $f'(g(x))=-sin(2x)$ .

Finally, if $g(x)=2x$ , its derivative $g'(x)$ is simply $2$ .

Now let's put the pieces together:

$n f^{n-1}(g(x)) * f'(g(x)) * g'(x)=$

$4 cos^3(2x) * (-sin(2x)) * 2=$

$-8sin(2x)cos^3(2x)$

How do you find the derivative of Cos^4(2x)Cos^4(2x)?