Derivative of a Trig Equation

[Denomination]

Find f'(x) when

f(x) = 3cos2x - 2(cosx)^2

I know the first part comes out to:

-6sin2x - 4(cosx)-sinx

Then I know that it would come out to:

-6(2sinxcosx) + 4 cosxsinx because sin2x= 2sinxcosx

which would lead to:

-12sinxcosx + 4 cosxsinx
then:
-8sinxcosx
and because they want the answer back in sin2x form it would be

-4(2sinxcosx)
and final answer be
-4sin2x
but I tried that as my answer for my online homework and it said the answer was incorrect. I think that I might be doing something wrong with that negative sin at the beginning, and if not it has to be bad algebra somewhere but I just can't find where. I would appreciate any kind of help. Thanks.

 

[tkhunny]

f(x) = 3cos2x - 2(cosx)^2
f'(x) = -6sin2x - 4(cosx)-sinx

Fix your notation.

f'(x) = -6sin(2x) - 4(cos(x))(-sin(x))

There is a price to pay for sloppy notation. Don't pay it. Fix it.

Why go through sin(x) and cos(x) if you know already that the required answer is in sin(2x)?

f'(x) = -6sin(2x) + 4(cos(x))(sin(x)) = -6sin(2x) + 2(2sin(x)cos(x)) = -6sin(2x) + 2sin(2x) = -4sin(2x)

We can verify this with a little different method.

f(x) = 3cos(2x) - 2(cos(x))^2 = 3(2(cos(x))^2 - 1) - 2(cos(x))^2 = 4(cos(x))^2 - 3

f'(x) = 4*(2)*cos(x)*(-sin(x)) = -4(2sin(x)cos(x)) = -4sin(2x)

Perhaps the "wrong" part would be the sloppy notation. Notice how I wrote the equation EVERY time. You have only expressions through most of it. Be more formal about it. The notation WILL help you.

 

[Denomination]

Thanks

Yeah that does seem a lot more logical. Thank You.