Another Trig Function with an Exponent

[Jason76]

\(\displaystyle y = \cos^{2} (2x)\)

\(\displaystyle y' = 2 \cos (2x) (-\sin 2x)(2) dx \)

\(\displaystyle y' = -4 \sin 2x \cos 2x dx \)

\(\displaystyle y' = -2 \sin 4x dx\) Everything makes sense, except this line.

 

[MarkFL]

I would do one of two things with your notation...either replace \(\displaystyle y'\) with \(\displaystyle dy\), or omit the \(\displaystyle dx\).

As far as the line you don't understand, what is the double-angle identity for sine?

 

[Jason76]

I would do one of two things with your notation...either replace \(\displaystyle y'\) with \(\displaystyle dy\), or omit the \(\displaystyle dx\).

As far as the line you don't understand, what is the double-angle identity for sine?

Double angle for sine: \(\displaystyle \sin(2x) = 2\sinx\cosx\) which we see in third line, as the 2x can be factored out to produce: \(\displaystyle 2\sinx\cosx\)