C3 Differentiation

cos^2(x)-sin^2(x)=cos(2x)

Unnecessary (and indeed not very useful), no?

cos^2(x)-sin^2(x)=cos(2x)

You have d/dx (1/2 sin x cos x) = 1/2 (cos^2 x - sin^2 x), and then have decided that that actually meant 1/2 (cos^2 x + sin^2 x). Sign error. Nothing major.

huh? I didn't write 1/2 (cos^2 x + sin^2 x) anywhere! And I don't think any of the signs are wrong.

If I use this, I get:
(1/2)cos(2x)
so how would I get to $sin^2 x$?

Its even easier to say:
2y=x-sinxcosx
2y=x-1/2sin2x
[sin2x]

Same to you as to nota: this might make it neater, but it makes it no easier to differentiate, and it just makes it less obvious how to proceed once you've differentiated.

Much as I sympathise, hitting everything with random trig identities isn't always the way to go.

$\frac{dy}{dx}$

$= \frac{1}{2} - ( \frac{1}{2}cos^2 x - \frac{1}{2}sin^2 x)$

$= \frac{1}{2} - \frac{1}{2}cos^2 x + \frac{1}{2}sin^2 x$

$= \frac{1}{2} (1 - cos^2 x + sin^2 x)$

$= \frac{1}{2} (1 - 1)$

$= \frac{1}{2} (0)$

$= 0$

Now I'm completely confused.

I'm kinda dumb *boink*

But anyways before you differentiate, you should use $\sin 2x = 2 \sin x \cos x$, and after you differentiate you should use $\sin^2 x=\frac{1-cos2x}{2}$ (which is just $\cos2x = 1 - 2 \sin^2 x$ rearranged).

2y=x - 1/2sin2x
2dy/dx=1-cos2x
dy/dx=1/2(1-cos2x) and from [cos2x] or using a standard result:
dy/dx=sin^2(x}

No product rule required...lol

No, but the chain rule's required, which is about as taxing as the product rule, and it means you have to apply the same trig identity twice, once forwards, once backwards. Seems silly to me.

No, but the chain rule's required, which is about as taxing as the product rule, and it means you have to apply the same trig identity twice, once forwards, once backwards. Seems silly to me.