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)#