# 1st year calc, simple trig derivative

1. May 25, 2006

### m0286

Hello!
This is probably real simple, just Im confused the question is Differentiate the following
y=sin^2 x - cos^2 x
I know you need to use just f'(x)+g'(x)
and I know the derivate of sinx is cosx and derivative of cosx is -sinx.... but what do you do with the sin^2.... is the derivate of sin^2x just cos^2x??? and derivate of cos^2x just -sin^2x? THANKS

2. May 25, 2006

### Staff: Mentor

An easy way to think about it is cos^2(x) = cos(x) * cos(x), and use the chain rule (first times the derivative of the second, plus the second times the derivitave of the first)...

3. May 26, 2006

### Pseudo Statistic

You could make it even easier by substituting in:
sin^2 x = 1 - cos^2 x
Then you get:
y = 1 - cos^2 x - cos^2 x = 1 - 2cos^2 x
And then you only have to use the chain rule (or product rule) once. :)

4. May 26, 2006

### finchie_88

Unless I'm mistaken, $cos^2 x - sin^2 x = cos 2x$, so can you use that?

Last edited by a moderator: May 26, 2006
5. May 26, 2006

### HallsofIvy

Staff Emeritus
Yes, it can be done that way, but that's not the "chain rule", it is the "product rule".

To use the chain rule, let u(x)= cos(x) and y(u)= u2. Then dy/dx= (dy/du)(du/dx). That will give the derivative of cos2(x).
Also, of course, if you let u(x)= sin(x) and y(u)= u2 you can do exactly the same thing to get the derivative of sin2(x).

I assume that you know that d(f(x)+ g(x))/dx= df/dx+ dg/dx.

Looks like there are 3 different ways to do this!

6. May 26, 2006

### Staff: Mentor

Oopsies, thanks for the correction.

7. May 27, 2006

### m0286

hello!
Thanks alot for the help... I decided to use the product rule since I am most familiar with that.
So once I did that I ended up with
(sinx)(cosx)+(sinx)(cosx) and (cosx)(-sinx)+(cosx)(-sinx)
so you add them together for f'(x)=h'(x)+g'(x)
now i ALWAYS mess these big things up... so heres what i did:
((sinx)(cosx)+(sinx)(cosx)) + ((cosx)(-sinx)+(cosx)(-sinx))
=2(sinxcosx)+2(cosx(-sinx))
=2sinx+2cosx+2cosx-2sinx
=0sinx+4cosx
=4cosx
Now that doesnt seem right can someone help me with what I am doing wrong? Or was the answer just 2(sinxcosx)+2(cosx(-sinx))... and I took it to far??? THANKS!

8. May 27, 2006

### VietDao29

There are some minor problems in your work.
The plus sign is wrong, you can take a look at the problem again, it's
y = sin2x - cos2x :)
How can you go from the former expression to latter one? Are you trying to say that
sin(x)cos(x) = sin(x) + cos(x)?
------------
Apart from that, everything looks perfect. :)

9. May 27, 2006

### m0286

Ok thnx :)
I see it was a stupid mistake...
I went from
=2(sinxcosx)+2(cosx(-sinx)) to
=2sinx+2cosx+2cosx-2sinx Because i multiplied them by the 2 infront of the brackets.. guess thats wrong! haha but thanks. I always get confused when you do certain things and dont. Thanks for all your help!

10. May 28, 2006

### HallsofIvy

Staff Emeritus
I don't mean to put you down but you really need to review your algebra. My experience is that people who cannot do algebra easily have a very hard time with calculus.

Surely you cannot believe that
2(sin x cos x)= 2sin x+ 2cos x!

Because you were differentiating sin2x- cos2x you should have 2sin x cos x- (-2sin x cosx)= 2sin x cos x+ 2sin x cos x.

Remember that finchie-88 pointed out that cos2x- sin2x= cos(2x)? Take the derivative of cos(2x) and compare it with the above.