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albear · Jun 16, 2007, 12:54 PM

I'm having a problem differentiating y=2sin(t)^2

I get
4sin(t)*2cos(t) but this is wrong any ideas on how to do it right

galactus · Jun 16, 2007, 01:04 PM

You're close. Use the chain rule. Taking $u=sint$

$2\cdot\frac{d}{dt}[sin^{2}t]=2\cdot\frac{d}{dt}[u^{2}]=2\cdot2u\frac{du}{dt}=2\cdot2sin(t)\cdot\frac{d}{ dt}[sin(t)]=2\cdot2sin(t)cos(t)=4sin(t)cos(t)$

albear · Jun 16, 2007, 01:10 PM

wait so what's v then and the mark scheme gets (-2sintcost)/(cosect^2) which then goesto -2sin(^3)tcost

albear · Jun 16, 2007, 01:35 PM

Hello

galactus · Jun 16, 2007, 01:45 PM

I don't know about the mark scheme(Is that an answer key?), but the derivative of

$2sin^{2}t$ is $4sin(t)cos(t)$

The derivative of $\frac{-sin^{4}(t)}{2}$ is $-2sin^{3}(t)cos(t)$

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What's v?

I don't know what you mean.

I must say, Albear, you're good at writing cryptic posts.;)

May I make a suggestion? You have been posting for awhile. How about learning some LaTex. It's easy and will make your posts more understandable.