# Homework Help: Differentiating a trig function

1. Oct 14, 2007

### maphco

1. The problem statement, all variables and given/known data
Differentiate using the Chain Rule:
$$y=\cos^2(\frac{x^2 + 2}{x^2 - 2})$$

2. Relevant equations

3. 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.

Last edited: Oct 14, 2007
2. Oct 14, 2007

### 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

3. Oct 14, 2007

### rocomath

double angle identity

$$2sinxcosx = sin2x$$

4. Oct 14, 2007

### 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

5. Oct 14, 2007

### 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

6. Oct 14, 2007

### maphco

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