Question

If y = cos^2 (2x^2) , find y'.

Answer 1

\[y=\cos ^2\left( 2x^2 \right),y \prime=2\cos 2x^2*(-\sin 2x^2)*4x=-4x \sin 4x^2\]

Answer 2

So it's 2 cos 2x?

Answer 3

*2x^2?
Sorry I'm very confused

Answer 4

Oh wait I see the rest of it Nevermind thank you

Answer 5

\[\frac{ d }{ dx }x^n=nx ^{n-1}\]
\[\frac{ d }{ dx }\cos x=-\sin x\]

Answer 6

instead of x here \[2x^2\]