Differentiating a trig function

[maphco]

Homework Statement

Differentiate using the Chain Rule:
y=\cos^2(\frac{x^2 + 2}{x^2 - 2})

Homework Equations

The Attempt at a Solution

y' = -2\cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2}) [\frac{2x(x^2 - 2) - (x^2 + 2)2x}{(x^2 - 2)^2}]
\mbox{derivative of cos is -sin so I brought the negative to the front}
y' = -2\cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2}) [\frac{2x(x^2 - 2 - x^2 - 2)}{(x^2 - 2)^2}]
y' = -2\cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2})[\frac{-8x}{(x^2 - 2)^2}]
y' = 16x(x^2 - 2)^{-2} \cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2})

However the worksheet says the answer is:
y' = 8x(x^2 - 2)^{-2} \sin(\frac{2x^2 + 4}{x^2 - 2})

What the *flat line* did I do wrong? My answer isn't even close.

 

[bob1182006]

they simplified using a formula for:
\sin(\alpha)\cos(\beta)=\frac{1}{2}\sin(\alpha-\beta)+\frac{1}{2}\sin(\alpha+\beta)

when you plug in your values for sin(a)cos(b) the first sin cancels out

 

[rocomath]

double angle identity

2sinxcosx = sin2x

 

[maphco]

Bob, we haven't been shown the half angle identities, which is what I assume that formula is used for. I know those formulae, but can't use them because we haven't been shown them formally :p.

Roco, that is genius! :D

 

[rocomath]

if you find that you're differentiating cosine or sine to the 2nd power, i usually differentiate and simplify using double-angle identity

also, when you have secx to some power, re-write the power and just add tanx at the end.

\sec^{2}x

\sec^{2}x\tan{x}

makes the writing a lil easier

 

[maphco]

Roco, thanks for the idea on simplifying the 2nd power part.