1st year calc, simple trig derivative

[m0286]

Hello!
This is probably real simple, just I am 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

 

[berkeman]

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)...

 

[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. :)

 

[finchie_88]

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

 

[HallsofIvy]

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)...

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!

 

[berkeman]

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

Oopsies, thanks for the correction.

 

[m0286]

hello!
Thanks a lot 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 here's 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 doesn't 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!

 

[VietDao29]

There are some minor problems in your work.

now i ALWAYS mess these big things up... so here's what i did:
((sinx)(cosx)+(sinx)(cosx)) + ((cosx)(-sinx)+(cosx)(-sinx))

The plus sign is wrong, you can take a look at the problem again, it's
y = sin²x - cos²x :)

=2(sinxcosx)+2(cosx(-sinx))
=2sinx+2cosx+2cosx-2sinx

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. :)

 

[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 that's wrong! haha but thanks. I always get confused when you do certain things and dont. Thanks for all your help!

 

[HallsofIvy]

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.