how do you differentiate 2cos(theta) - 2cos^2(theta)?

i got -2sin(theta) +2sin^2(theta) but its wrong.....

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- Apr 9th 2007, 06:10 PMjephdifferentiate
how do you differentiate 2cos(theta) - 2cos^2(theta)?

i got -2sin(theta) +2sin^2(theta) but its wrong..... - Apr 9th 2007, 06:20 PMqbkr21Re:
Hi Jeph!!

Remember that:

y = cos(x) y' = -sin(x)

also

y = sin(x) y' = cos(x)

the prefix -co in this since means negative

http://item.slide.com/r/1/39/i/Xnn41...kuqNw7Aa7VcGv/ - Apr 9th 2007, 06:37 PMjeph
i tohught if you differentiate cos it would turn to sin not cos and sin...

can you do 2sin(theta) - 2cos(theta)sin(theta) for me? - Apr 9th 2007, 06:48 PMqbkr21Re:
Sure, but remember...

Cos(x) does not turn into Sin(x)

it turns into -Sin(x)

It's crucial that you put a negative in front of -Sin(x) or else your answer is no good. Also if there was a coefficient in front of Cos(x) such as 4Cos(x) the negative would go out in front thus -4Sin(x).

You will see some product rule action in this next one:

http://item.slide.com/r/1/43/i/Kcu-i...xfWe5qt7K8Pok/ - Apr 10th 2007, 06:59 AMJhevon
what happened was he differentiated cos

**^2**(theta), for this you would get sin(theta)cos(theta)

explanation:

we have to do the chain rule on this one.

it might be better to think of cos^2(theta) as (cos(theta))^2. that way, to perform the chain rule, you first have to bring the 2 down and multiply, subtract 1 from the power and leave what's in the brackets, then to complete the chain rule, multiply by the derivative of what's in the brackets.

y = cos^2(theta)

=> y = (cos(theta))^2

=> y' = 2(cos(theta))*(-sin(theta))

=> y' = -2cos(theta)sin(theta) - Apr 10th 2007, 11:53 AMjeph
oh...hmmmm

thanks guys. its slowly coming back to me alittle now...